Stoichiometry with Reactions - AP Chemistry

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Question

Octane, $C_{8}H_{18}$ , is an important of component of gasoline, which can be burned as fuel in cars. The following is an unbalanced equation for the combustion of octane.

If 114 g of octane are burned in the complete combustion reaction shown below, how many grams of water will be produced? $C_{8}H_{18} +$ $O_{2} \rightarrow$ $CO_{2}$ $H_{2}O$

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Answer

First, the equation should be balanced. This can be done with the coefficients 2, 25, 16, and 18.

Next, calculating the molar mass of octane shows that it has a molar mass of 114 g/mol. This means that 1 mole of octane is used in the reaction, and 9 times as many moles of water should be produced, meaning one of the products will be 9 mol of water. Water has a molar mass of 18 g/mol, so the total mass of water produced is $9\cdot18=162\textup g$ .

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Question

Suppose that 4 moles of undergoes complete decomposition. At STP, how many liters of would be produced as product? (At STP, one mole of a gas takes up 22.4 L of space.)

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Answer

Start by creating a skeleton equation for the decomposition reaction:

$NaClO_3 \rightarrow NaCl + O_2$

This can be balanced to give:

$2NaClO_3 \rightarrow 2NaCl + 3O_2$

This means that 4 mol of reactant would produce 6 mol of oxygen gas. The question states that at STP, one mole of a gas takes up 22.4 L of space, so it is easy to find that 6 moles of oxygen gas would occupy a volume of 134.4 L:

$6\text{ mol O}\times\frac{22.4 L}{1 \text{ mol}}=134.4\text{ L}$

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Question

What volume of 0.5M NaOH is necessary to neutralize 15mL of 1.0M nitrous acid?

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Answer

Both the acid and the base provide one equivalent of the necessary ions (H+ and OH–, respectively) and therefore you can use the equation M1V1 = M2V2.

(1)(15) = (0.5)V2

V2 = 30mL

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Question

What volume of 0.5M Mg(OH)2 is necessary to neutralize 15mL of 1.0M nitrous acid?

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Answer

This problem is tricky because you must realize that Mg(OH)2 will release 2 molecules of OH– whenever one dissolves; thus, you only need half the amount of Mg(OH)2 to neutralize all of the nitrous acid, which only gives 1 equivalent.

M1V1 = M2V2

Plugging in the values, you get 30mL, but in reality you only need half of that amount.

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Question

Consider the reaction of baking soda and vinegar in an aqueous solution:

$NaHCO_{3\hspace{1 mm}(s)}+CH_3 COOH_{(l)}\rightarrow CO_{2\hspace{1 mm}(g)}+H_2O_{(l)}+Na^{+}{(aq)}+CH_3COO^{-}{(aq)}$

If we begin with 17g of $NaHCO_3$ and excess $CH_3COOH$ , what is the volume of the gas released?

Assume that all gases behave as ideal gases at atmospheric pressure. The temperature is 298K.

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Answer

We begin with 17 g of $NaHCO_3$ , so our first step is to calculate the molecular weight of $NaHCO_3$ .

$\frac{23.0\hspace{1 mm} g\hspace{1 mm}Na}{1\hspace{1 mm}mole\hspace{1 mm}Na}+\frac{1.0\hspace{1 mm}g\hspace{1 mm}H}{1\hspace{1 mm}mole\hspace{1 mm}H}+\frac{12.0\hspace{1 mm}g\hspace{1 mm}C}{1\hspace{1 mm}mole\hspace{1 mm}C}+3\times(\frac{16.0\hspace{1 mm}g\hspace{1 mm}O}{1\hspace{1 mm}mole\hspace{1 mm}O})=\frac{84.0\hspace{1 mm}g\hspace{1 mm}NaHCO_3}{1\hspace{1 mm}mole\hspace{1 mm}NaHCO_3}$

Now with the molecular weight, we will calculate the number of moles within 17 g of $NaHCO_3$ .

$17\hspace{1 mm}g\hspace{1 mm}NaHCO_3\times \frac{1\hspace{1 mm}mole\hspace{1 mm}NaHCO_3}{84.0\hspace{1 mm}g\hspace{1 mm}NaHCO_3}=2.03\times10^{-1}mole\hspace{1 mm}NaHCO_3$

Now we will use this to calculate the number of moles of gas produced. In this reaction, the only gas produced is carbon dioxide.

$2.03\times10^{-1}mole\hspace{1 mm}NaHCO_3\hspace{1 mm}\times\frac{1\hspace{1 mm}mole\hspace{1 mm}CO_2}{1\hspace{1 mm}mole\hspace{1 mm}NaHCO_3}=2.03\times10^{-1}moles\hspace{1 mm}CO_2$

An ideal gas at 1 atmosphere and 298 K has the molar volume of $24.465 \frac{L}{mol}$ .

$2.03\times10^{-1}\hspace{1 mm}mol\hspace{1 mm}CO_2\times \frac{24.465\hspace{1 mm}L\hspace{1 mm}gas}{1\hspace{1 mm}mol\hspace{1 mm}gas}=4.95\hspace{1 mm}L\hspace{1 mm}CO_2$

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Question

Consider the following chemical reaction:

$Al_2O_3(s)+3H_2SO_4(aq)\rightarrow Al_2(SO_4)_3(aq)+3H_2O(l)$

How many moles of will be produced if 75.0mL of water is produced?

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Answer

$75.0\hspace{1 mm}mL\times\frac{1\hspace{1 mm}g\hspace{1 mm}H_2O}{1\hspace{1 mm}mL\hspace{1 mm}H_2O}\times\frac{1\hspace{1 mm}mole\hspace{1 mm}H_2O}{18.0\hspace{1 mm}g\hspace{1 mm}H_2O}\times\frac{1\hspace{1 mm}mole\hspace{1 mm}Al_2(SO_4)_3}{3\hspace{1 mm}moles\hspace{1 mm}H_2O}=1.39\hspace{1 mm}moles\hspace{1 mm}Al_2(SO_4)_3$

This problem is solved based on stoichiometry. Using the molecular weights and ratios in the given reaction, we can solve for the amount of produced.

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Question

What volume of chlorine gas at is needed to react with 14.2g of sodium to form NaCl at 1.72atm?

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Answer

$2Na(s)+Cl_2(g)\rightarrow 2NaCl(s)$

First, we need to solve for the moles of chlorine gas needed to react with the full amount of sodium metal.

$14.2g\ Na\frac{1mol\ Na}{23g\ Na}\frac{1mol\ Cl_2}{2mol\ Na}=0.31mol\ Cl_2$

Now, using the ideal gas law, we can solve for the volume that this gas would occupy at the given temperature and pressure.

$V=\frac{nRT}{p}$

$V=\frac{(0.31mol\ Cl_2)(0.0821\frac{L\cdot atm}{mol\cdot K})(45.3^oC)}{1.72atm}$

$V=\frac{(0.31mol\ Cl_2)(0.0821\frac{L\cdot atm}{mol\cdot K})(318.3K)}{1.72atm}$

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Question

The formation of ammonia is given by the following equation:

$3H_2+N_2\rightarrow 2NH_3$

Assuming you start with 15mol of hydrogen gas, how many moles of nitrogen gas will be needed to completely react the hydrogen gas?

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Answer

Since the amount of hydrogen gas is given in moles, this solution is as easy as using the molar ratio between the reactants. Looking at the equation, we see that there is a 1:3 ratio for nitrogen to hydrogen. As a result, we would need one third of the amount of hydrogen available in order to use up all of the hydrogen gas.

$15mol H_2*\frac{1molN_2}{3mol H_2}=5molN_2$

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Question

The formation of ammonia is given by the following equation:

$3H_2+N_2\rightarrow 2NH_3$

Assuming you started with 20g of hydrogen gas, how much nitrogen gas is necessary to react the hydrogen gas?

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Answer

We can convert the mass of hydrogen from grams to moles by using the equation moles = mass / molar mass

$\frac{20gH_2}{2\frac{g}{mol}H_2}=10mol H_2$

Now that we have the moles of hydrogen, we can use the molar ratio given in the equation of 1:3. Since we need 1/3 of the moles of hydrogen as nitrogen, we know we need 3.33 moles of nitrogen gas.

$10molH_2\frac{1molN_2}{3molH_2}=3.33molN_2$

Now, we simply multiply the moles by the molar mass of nitrogen gas (28 grams per mol) and we find we need 93.33 grams of nitrogen gas.

$3.33molN_228\frac{g}{mol}N_2=93.33gN_2$

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Question

Magnesium will combine with oxygen gas to form magnesium oxide according to the balanced equation below.

$2Mg + O_{2}\rightarrow2MgO$

If 65g of MgO are created in the reaction, how many grams of magnesium are used in the reaction?

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Answer

In order to determine how many grams of magnesium are used in the reaction, we can use the percentage by mass of magnesium in MgO in order to find out how much magnesium is in 65 grams of MgO. The molar mass of MgO is 40.3 grams. The molar mass of magnesium is 24.3 grams. Dividing 24.3 by 40.3 reveals that magnesium makes up 60% of the mass of MgO.

$\frac{24.3g\ Mg}{40.3g\ MgO}=0.603\sim 60%$

Multiplying 65 grams by 0.603 reveals that there are 39.2 grams of magnesium in 65 grams of MgO.

$65.0g\ MgO*0.603=39.2g\ Mg$

An alternative means of solving the problem would be to convert grams of MgO to moles. The ratio of MgO to Mg in the given chemical equation is 2:2, thus the number of moles of MgO formed is equal to the number of moles of Mg reacted. Once this number is found, it can be converted back to grams using the molar mass of magnesium.

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Question

How many moles of water vapor will be formed when 5.00g of hydrogen gas react with excess oxygen?

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Answer

When dealing with questions related to stoichiometry, it is important to compare only mole-to-mole ratios.

Hydrogen gas is diatomic. When converting grams to moles, we will need to double the atomic mass of hydrogen.

$5g\ H_2\frac{1mol\ H_2}{2.0g\ H_2}=2.5mol\ H_2$

We now know we have 2.5 moles of hydrogen gas. Now we need to write a balanced equation to determine the conversion factor between hydrogen gas and the product.

$2H_2(g)+O_2(g)\rightleftharpoons 2H_2O(g)$

For every two moles of hydrogen consumed, two moles of water are produced. Using this ratio and the starting moles of hydrogen, we can calculate the moles of water produced.

$2.5mol\ H_2\frac{2mol\ H_2O}{2mol\ H_2}=2.5mol\ H_2O$

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Question

What volume of 0.5M NaOH is necessary to neutralize 15mL of 1.0M nitrous acid?

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Answer

Both the acid and the base provide one equivalent of the necessary ions (H+ and OH–, respectively) and therefore you can use the equation M1V1 = M2V2.

(1)(15) = (0.5)V2

V2 = 30mL

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Question

What volume of 0.5M Mg(OH)2 is necessary to neutralize 15mL of 1.0M nitrous acid?

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Answer

This problem is tricky because you must realize that Mg(OH)2 will release 2 molecules of OH– whenever one dissolves; thus, you only need half the amount of Mg(OH)2 to neutralize all of the nitrous acid, which only gives 1 equivalent.

M1V1 = M2V2

Plugging in the values, you get 30mL, but in reality you only need half of that amount.

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Question

Consider the reaction of baking soda and vinegar in an aqueous solution:

$NaHCO_{3\hspace{1 mm}(s)}+CH_3 COOH_{(l)}\rightarrow CO_{2\hspace{1 mm}(g)}+H_2O_{(l)}+Na^{+}{(aq)}+CH_3COO^{-}{(aq)}$

If we begin with 17g of $NaHCO_3$ and excess $CH_3COOH$ , what is the volume of the gas released?

Assume that all gases behave as ideal gases at atmospheric pressure. The temperature is 298K.

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Answer

We begin with 17 g of $NaHCO_3$ , so our first step is to calculate the molecular weight of $NaHCO_3$ .

$\frac{23.0\hspace{1 mm} g\hspace{1 mm}Na}{1\hspace{1 mm}mole\hspace{1 mm}Na}+\frac{1.0\hspace{1 mm}g\hspace{1 mm}H}{1\hspace{1 mm}mole\hspace{1 mm}H}+\frac{12.0\hspace{1 mm}g\hspace{1 mm}C}{1\hspace{1 mm}mole\hspace{1 mm}C}+3\times(\frac{16.0\hspace{1 mm}g\hspace{1 mm}O}{1\hspace{1 mm}mole\hspace{1 mm}O})=\frac{84.0\hspace{1 mm}g\hspace{1 mm}NaHCO_3}{1\hspace{1 mm}mole\hspace{1 mm}NaHCO_3}$

Now with the molecular weight, we will calculate the number of moles within 17 g of $NaHCO_3$ .

$17\hspace{1 mm}g\hspace{1 mm}NaHCO_3\times \frac{1\hspace{1 mm}mole\hspace{1 mm}NaHCO_3}{84.0\hspace{1 mm}g\hspace{1 mm}NaHCO_3}=2.03\times10^{-1}mole\hspace{1 mm}NaHCO_3$

Now we will use this to calculate the number of moles of gas produced. In this reaction, the only gas produced is carbon dioxide.

$2.03\times10^{-1}mole\hspace{1 mm}NaHCO_3\hspace{1 mm}\times\frac{1\hspace{1 mm}mole\hspace{1 mm}CO_2}{1\hspace{1 mm}mole\hspace{1 mm}NaHCO_3}=2.03\times10^{-1}moles\hspace{1 mm}CO_2$

An ideal gas at 1 atmosphere and 298 K has the molar volume of $24.465 \frac{L}{mol}$ .

$2.03\times10^{-1}\hspace{1 mm}mol\hspace{1 mm}CO_2\times \frac{24.465\hspace{1 mm}L\hspace{1 mm}gas}{1\hspace{1 mm}mol\hspace{1 mm}gas}=4.95\hspace{1 mm}L\hspace{1 mm}CO_2$

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Question

Consider the following chemical reaction:

$Al_2O_3(s)+3H_2SO_4(aq)\rightarrow Al_2(SO_4)_3(aq)+3H_2O(l)$

How many moles of will be produced if 75.0mL of water is produced?

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Answer

$75.0\hspace{1 mm}mL\times\frac{1\hspace{1 mm}g\hspace{1 mm}H_2O}{1\hspace{1 mm}mL\hspace{1 mm}H_2O}\times\frac{1\hspace{1 mm}mole\hspace{1 mm}H_2O}{18.0\hspace{1 mm}g\hspace{1 mm}H_2O}\times\frac{1\hspace{1 mm}mole\hspace{1 mm}Al_2(SO_4)_3}{3\hspace{1 mm}moles\hspace{1 mm}H_2O}=1.39\hspace{1 mm}moles\hspace{1 mm}Al_2(SO_4)_3$

This problem is solved based on stoichiometry. Using the molecular weights and ratios in the given reaction, we can solve for the amount of produced.

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Question

What volume of chlorine gas at is needed to react with 14.2g of sodium to form NaCl at 1.72atm?

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Answer

$2Na(s)+Cl_2(g)\rightarrow 2NaCl(s)$

First, we need to solve for the moles of chlorine gas needed to react with the full amount of sodium metal.

$14.2g\ Na\frac{1mol\ Na}{23g\ Na}\frac{1mol\ Cl_2}{2mol\ Na}=0.31mol\ Cl_2$

Now, using the ideal gas law, we can solve for the volume that this gas would occupy at the given temperature and pressure.

$V=\frac{nRT}{p}$

$V=\frac{(0.31mol\ Cl_2)(0.0821\frac{L\cdot atm}{mol\cdot K})(45.3^oC)}{1.72atm}$

$V=\frac{(0.31mol\ Cl_2)(0.0821\frac{L\cdot atm}{mol\cdot K})(318.3K)}{1.72atm}$

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Question

The formation of ammonia is given by the following equation:

$3H_2+N_2\rightarrow 2NH_3$

Assuming you start with 15mol of hydrogen gas, how many moles of nitrogen gas will be needed to completely react the hydrogen gas?

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Answer

Since the amount of hydrogen gas is given in moles, this solution is as easy as using the molar ratio between the reactants. Looking at the equation, we see that there is a 1:3 ratio for nitrogen to hydrogen. As a result, we would need one third of the amount of hydrogen available in order to use up all of the hydrogen gas.

$15mol H_2*\frac{1molN_2}{3mol H_2}=5molN_2$

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Question

The formation of ammonia is given by the following equation:

$3H_2+N_2\rightarrow 2NH_3$

Assuming you started with 20g of hydrogen gas, how much nitrogen gas is necessary to react the hydrogen gas?

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Answer

We can convert the mass of hydrogen from grams to moles by using the equation moles = mass / molar mass

$\frac{20gH_2}{2\frac{g}{mol}H_2}=10mol H_2$

Now that we have the moles of hydrogen, we can use the molar ratio given in the equation of 1:3. Since we need 1/3 of the moles of hydrogen as nitrogen, we know we need 3.33 moles of nitrogen gas.

$10molH_2\frac{1molN_2}{3molH_2}=3.33molN_2$

Now, we simply multiply the moles by the molar mass of nitrogen gas (28 grams per mol) and we find we need 93.33 grams of nitrogen gas.

$3.33molN_228\frac{g}{mol}N_2=93.33gN_2$

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Question

Magnesium will combine with oxygen gas to form magnesium oxide according to the balanced equation below.

$2Mg + O_{2}\rightarrow2MgO$

If 65g of MgO are created in the reaction, how many grams of magnesium are used in the reaction?

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Answer

In order to determine how many grams of magnesium are used in the reaction, we can use the percentage by mass of magnesium in MgO in order to find out how much magnesium is in 65 grams of MgO. The molar mass of MgO is 40.3 grams. The molar mass of magnesium is 24.3 grams. Dividing 24.3 by 40.3 reveals that magnesium makes up 60% of the mass of MgO.

$\frac{24.3g\ Mg}{40.3g\ MgO}=0.603\sim 60%$

Multiplying 65 grams by 0.603 reveals that there are 39.2 grams of magnesium in 65 grams of MgO.

$65.0g\ MgO*0.603=39.2g\ Mg$

An alternative means of solving the problem would be to convert grams of MgO to moles. The ratio of MgO to Mg in the given chemical equation is 2:2, thus the number of moles of MgO formed is equal to the number of moles of Mg reacted. Once this number is found, it can be converted back to grams using the molar mass of magnesium.

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Question

How many moles of water vapor will be formed when 5.00g of hydrogen gas react with excess oxygen?

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Answer

When dealing with questions related to stoichiometry, it is important to compare only mole-to-mole ratios.

Hydrogen gas is diatomic. When converting grams to moles, we will need to double the atomic mass of hydrogen.

$5g\ H_2\frac{1mol\ H_2}{2.0g\ H_2}=2.5mol\ H_2$

We now know we have 2.5 moles of hydrogen gas. Now we need to write a balanced equation to determine the conversion factor between hydrogen gas and the product.

$2H_2(g)+O_2(g)\rightleftharpoons 2H_2O(g)$

For every two moles of hydrogen consumed, two moles of water are produced. Using this ratio and the starting moles of hydrogen, we can calculate the moles of water produced.

$2.5mol\ H_2\frac{2mol\ H_2O}{2mol\ H_2}=2.5mol\ H_2O$

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