Balancing Reactions - AP Chemistry

Card 0 of 196

Question

When balanced, what is the value of in the following chemical equation?

$\text{SO}_2+x\text{Li}_2\text{Se}\rightarrow\text{SSe}_2+\text{Li}_2\text{O}$

Answer

Recall that a balanced chemical equation has the same number of each element on one side.

$\text{SO}_2+x\text{Li}_2\text{Se}\rightarrow\text{SSe}_2+\text{Li}_2\text{O}$

Start by counting the number of each element on each side.

There are the following numbers of moles of each reactant:

$\text{Sulfur: 1}$

$\text{Oxygen: 2}$

$\text{Lithium: 2}$

$\text{Selenium: 1}$

There are the following numbers of moles of each product:

$\text{Sulfur: 1}$

$\text{Oxygen: 1}$

$\text{Lithium: 2}$

$\text{Selenium: 2}$

Add coefficients in front of the molecular compounds in the equation until there are the same numbers of sulfur, oxygen, lithium, and selenium on each side.

The balanced chemical equation is the following:

$\text{SO}_2+2\text{Li}_2\text{Se}\rightarrow\text{SSe}_2+2\text{Li}_2\text{O}$

Both products and reactants now have the following number of moles:

$\text{Sulfur: 1}$

$\text{Oxygen: 2}$

$\text{Lithium: 4}$

$\text{Selenium: 2}$

Since the coefficient in front of $\text{Li}_2\text{Se}$ is , must equal to .

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Question

Balance the following chemical equation:

$\text{Pb(NO}_3)_2(aq)+\text{KCl}(aq)\rightarrow \text{PbCl}_2(s)+\text{KNO}_3(aq)$

Answer

A balanced chemical equation will have the same number of each atom on both sides of the reaction.

$\text{Pb(NO}_3)_2(aq)+\text{KCl}(aq)\rightarrow \text{PbCl}_2(s)+\text{KNO}_3(aq)$

Start by balancing the number of nitrate.

Since we have nitrate on the left, we must also have nitrate on the right.

$\text{Pb(NO}_3)_2(aq)+\text{KCl}(aq)\rightarrow \text{PbCl}_2(s)+2\text{KNO}_3(aq)$

This equation now gives potassium on the left, and potassium on the right. Balance the potassium.

$\text{Pb(NO}_3)_2(aq)+2\text{KCl}(aq)\rightarrow \text{PbCl}_2(s)+2\text{KNO}_3(aq)$

This equation is balanced because there are equal numbers of each atom on both sides.

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Question

The following ReDox reaction takes place in acidic solution:

Fe2+ + Cr2O72– → Fe3+ + Cr3+

What is the sum of coefficients in this redox reaction?

Answer

When you balance the redox reaction in acidic conditons, there are 6Fe2+, 1 Cr2O72–, 14 H+, 6 Fe3+, 2 Cr3+, and 7 H2O. Don't forget to add the 1 in front of the Cr2O72–

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Question

For the following unbalanced redox reaction, how many electrons are transferred and which chemical species is being oxidized?

$PF_2I_{(l)}+Hg_{(l)}\rightarrow P_2F_4_{(g)}+Hg_2I_2_{(s)}$

Answer

To begin, we will need to separate the given reaction into the two half-reactions by identifying changes in oxidation number. In this case, mercury (Hg) and phosphorus (P) show a change in oxidation number. Mercury begins with an oxidation number of zero, and ends with an oxidation number of . Phosphorus begins with an oxidation number of and ends with an oxidation number of . Note that the oxidation numbers for fluorine and iodine reamain constant at for each.

Now we can begin to look at the half-reactions.

$PF_2I_{(l)}\rightarrow P_2F_4_{(g)}$

$Hg_{(l)}\rightarrow Hg_2I_2_{(s)}$

Balance the atoms.

$2PF_2I_{(l)}\rightarrow P_2F_4_{(g)}$

$2Hg_{(l)}\rightarrow Hg_2I_2_{(s)}$

Now balance the electrons. We know that each mercury atom loses one electron and each fluorine atom gains one electron.

$2PF_2I_{(l)}+2e^-\rightarrow P_2F_4_{(g)}$

$2Hg \rightarrow Hg_2I_2_{(s)}+2e^{-}$

We can see that two electrons are tranferred. To identify the element being oxidized, we must find the element that is losing electrons. In this case, mercury is being oxidized.

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Question

How many electrons are involved in the following reaction?

$MnO_4^-+ I^-\rightarrow I_2+Mn^{2+}$

Answer

The common factor between 2 e- and 5 e- is 10. Therefore the number of electrons involved is 10 e-.

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Question

How many electrons are involved in the following reaction?

$MnO_4^-+ Fe^{2+} \rightarrow Fe^{3+}+Mn^{2+}$

Answer

The common factor between 1 e- and 5 e- is 5. Therefore the number of electrons involved is 5 e-.

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Question

What is the balanced coefficient on OH- for the following reaction:

$Br^- + MnO_4^- \rightarrow MnO_2 + BrO_3^-$ (under basic conditions)

Answer

Add them together:

$3 H_2 O + Br^- + 2 MnO_4^- + 8 H^+ + 6 e^- \rightarrow 2 MnO_2 + BrO_3^- + 6 H^+ + 4 H_2 O + 6 e^-$

Simplify:

$Br^- + 2 MnO_4^- + 2 H^+ \rightarrow 2 MnO_2 + BrO_3^- + H_2 O$

Add Hydroxides to each side to counter H+.

$Br^- + 2 MnO_4^- + 2 H_2 O \rightarrow 2 MnO_2 + BrO_3^- + H_2 O+2 OH^-$

Simplify:

$Br^- + 2 MnO_4^- + H_2 O \rightarrow 2 MnO_2 + BrO_3^- +2 OH^-$

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Question

What is the sum of all the balanced coefficients in the following reaction:

$O_2 + Sb \rightarrow H_2 O_2 + SbO_2^-$ (basic conditions)

Answer

Add the equations together

$2Sb+4H_2 O + 3 O_2 + 6 H^+ + 6e^- \rightarrow 2SbO_2^-+8H^++3 H_2 O_2 + 6e^-$

Simplify

$2Sb+4H_2 O+3 O_2 \rightarrow 2SbO_2^-+2H^++3H_2 O_2$

Add 2 OH- to each side to cancel out the H+.

$2Sb+4H_2 O+3 O_2 + 2 OH^- \rightarrow 2SbO_2^-+2H_2 O+3H_2 O_2$

Simplify:

$2Sb+2 H_2 O+3 O_2 + 2 OH^- \rightarrow 2SbO_2^-+3H_2 O_2$

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Question

The following ReDox reaction takes place in acidic solution:

Fe2+ + Cr2O72– → Fe3+ + Cr3+

What is the sum of coefficients in this redox reaction?

Answer

When you balance the redox reaction in acidic conditons, there are 6Fe2+, 1 Cr2O72–, 14 H+, 6 Fe3+, 2 Cr3+, and 7 H2O. Don't forget to add the 1 in front of the Cr2O72–

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Question

For the following unbalanced redox reaction, how many electrons are transferred and which chemical species is being oxidized?

$PF_2I_{(l)}+Hg_{(l)}\rightarrow P_2F_4_{(g)}+Hg_2I_2_{(s)}$

Answer

To begin, we will need to separate the given reaction into the two half-reactions by identifying changes in oxidation number. In this case, mercury (Hg) and phosphorus (P) show a change in oxidation number. Mercury begins with an oxidation number of zero, and ends with an oxidation number of . Phosphorus begins with an oxidation number of and ends with an oxidation number of . Note that the oxidation numbers for fluorine and iodine reamain constant at for each.

Now we can begin to look at the half-reactions.

$PF_2I_{(l)}\rightarrow P_2F_4_{(g)}$

$Hg_{(l)}\rightarrow Hg_2I_2_{(s)}$

Balance the atoms.

$2PF_2I_{(l)}\rightarrow P_2F_4_{(g)}$

$2Hg_{(l)}\rightarrow Hg_2I_2_{(s)}$

Now balance the electrons. We know that each mercury atom loses one electron and each fluorine atom gains one electron.

$2PF_2I_{(l)}+2e^-\rightarrow P_2F_4_{(g)}$

$2Hg \rightarrow Hg_2I_2_{(s)}+2e^{-}$

We can see that two electrons are tranferred. To identify the element being oxidized, we must find the element that is losing electrons. In this case, mercury is being oxidized.

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Question

How many electrons are involved in the following reaction?

$MnO_4^-+ I^-\rightarrow I_2+Mn^{2+}$

Answer

The common factor between 2 e- and 5 e- is 10. Therefore the number of electrons involved is 10 e-.

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Question

How many electrons are involved in the following reaction?

$MnO_4^-+ Fe^{2+} \rightarrow Fe^{3+}+Mn^{2+}$

Answer

The common factor between 1 e- and 5 e- is 5. Therefore the number of electrons involved is 5 e-.

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Question

What is the balanced coefficient on OH- for the following reaction:

$Br^- + MnO_4^- \rightarrow MnO_2 + BrO_3^-$ (under basic conditions)

Answer

Add them together:

$3 H_2 O + Br^- + 2 MnO_4^- + 8 H^+ + 6 e^- \rightarrow 2 MnO_2 + BrO_3^- + 6 H^+ + 4 H_2 O + 6 e^-$

Simplify:

$Br^- + 2 MnO_4^- + 2 H^+ \rightarrow 2 MnO_2 + BrO_3^- + H_2 O$

Add Hydroxides to each side to counter H+.

$Br^- + 2 MnO_4^- + 2 H_2 O \rightarrow 2 MnO_2 + BrO_3^- + H_2 O+2 OH^-$

Simplify:

$Br^- + 2 MnO_4^- + H_2 O \rightarrow 2 MnO_2 + BrO_3^- +2 OH^-$

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Question

What is the sum of all the balanced coefficients in the following reaction:

$O_2 + Sb \rightarrow H_2 O_2 + SbO_2^-$ (basic conditions)

Answer

Add the equations together

$2Sb+4H_2 O + 3 O_2 + 6 H^+ + 6e^- \rightarrow 2SbO_2^-+8H^++3 H_2 O_2 + 6e^-$

Simplify

$2Sb+4H_2 O+3 O_2 \rightarrow 2SbO_2^-+2H^++3H_2 O_2$

Add 2 OH- to each side to cancel out the H+.

$2Sb+4H_2 O+3 O_2 + 2 OH^- \rightarrow 2SbO_2^-+2H_2 O+3H_2 O_2$

Simplify:

$2Sb+2 H_2 O+3 O_2 + 2 OH^- \rightarrow 2SbO_2^-+3H_2 O_2$

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Question

_Fe2O3 + _HCl ⇌ _FeCl3 + _H2O

The Following question will be based on the unbalanced reaction above

For the reaction above to be balanced what coefficient should be in front of the compound HCl?

Answer

Balancing reactions is best acheived by using a stepwise approach. It is useful to work through each atom making sure it is present in a balanced fashion on both sides of the equation. Starting with Fe, Fe2O3 is the only molecule with Fe present on the left side of the equation and FeCl3 is the only molecule with Fe present on the right side of the equation. Thus the molar ratio of Fe2O3 to FeCl3 must be 1:2. Moving on to O, the O on the left side of the equation exists as Fe2O3 and H2O on the right. The molar ratio of Fe2O3 to H2O is therefore 1:3. Cl exists in HCl on the left and FeCl3 on the right. Thus the molar ratio of HCl to FeCl3 must be 3:1. To ensure that these molar ratios are maintaned the balanced formula is then determined to be Fe2O3 + 6HCl -> 2FeCl3 + 3H2O

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Question

Balance the following equation:

FeCl3 + NOH5 → Fe(OH)3 + NH4Cl

Answer

The first thing to do is to balance the Cl (3 on the left; one on the right) ← add 3 for NH4Cl

Now, there are 3 N on the right and only one on the left; add a 3 to the NOH5

Check to see that the Fe, H, and O balance, which they do

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Question

After the following reaction is balanced, how many moles of H2O can 4 moles of C2H6 produce?

__C2H6(s) + O2(g) → H2O(l) + CO2(g)

Answer

The reaction balances out to 2C2H6(s) + 7O2(g) → 6H2O(l) + 4CO2(g). For every 2 moles of C2H6 you can produce 6 moles of H2O. Giving you a total production of 12 moles when you have 4 moles of C2H6.

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Question

Consider the following unbalanced equation for the combustion of propane, $C_3H_8$ :

$C_3H_{8\hspace{1 mm}(g)}+O_{2\hspace{1 mm}(g)}\rightarrow CO_{2\hspace{1 mm}(g)}+H_2O_{(l)}$

If you were to combust one mole of propane, how many moles of water would you produce?

Answer

Begin by balancing the equation. There are many ways to do this, but one method that is particularly useful is to assume that you have 1 mole of your hydrocarbon (propane), and balance the equation from there. It may be necessary to manipulate an equation further if you end up with fractions, but all you will need to do is multiply by an integer if that is the case.

First, we will balance the carbons. There are three carbons in propane, so we will make sure there are three carbons on the right side of the arrow as well:

$C_3H_8 + O_2 \rightarrow 3CO_2 + H_2O$

This is not complete. We will next balance the hydrogens. There are eight hydrogens on the left side of the equation, so:

$C_3H_8 + O_2 \rightarrow 3CO_2 + 4H_2O$

The last step is to balance the oxygens on the left and right side of the equation

$C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O$

Our equation is balanced and all coefficients are integers. If we begin with one mole of propane, we will produce four moles of water.

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Question

What is the chemical formula of the salt formed when a chemist mixes solvated Potassium and Arsenic ions in solution?

Answer

Potassium is a Group I element, so to get to a filled valence shell, it will lost one electron, yielding $K^+$ .

Arsenic is a Group 5 element, so it needs to gain three electrons to obtain a filled valence shell, yielding $As^{-3}$ .

In order to balance out the charges, the resultant salt will be $K_3As$ .

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Question

Consider the following unbalanced equation:

$Fe+Cl_2\rightarrow FeCl_3$

How many grams of solid iron are needed to make 36.0g of $FeCl_3$ ? Assume that chlorine is in excess.

Answer

$Fe+Cl_2\rightarrow FeCl_3$

First, we will balance the equation:

$2Fe+3Cl_2\rightarrow 2FeCl_3$

Since chlorine is in excess, we know that the limiting reagent is iron.

$36.0\hspace{1 mm}g\hspace{1 mm}FeCl_3\times\frac{1\hspace{1 mm}mole\hspace{1 mm}FeCl_3}{162.2\hspace{1 mm}g\hspace{1 mm}FeCl_3}\times\frac{2\hspace{1 mm}moles\hspace{1 mm}Fe}{2\hspace{1 mm}moles\hspace{1 mm}FeCl_3}\times\frac{55.845\hspace{1 mm}g\hspace{1 mm}Fe}{1\hspace{1 mm}mole\hspace{1 mm}Fe}=12.4\hspace{1 mm}g\hspace{1 mm}Fe$

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