AP Chemistry - Balancing Reactions

How many electrons are involved in the following reaction?

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

1 e-

2 e-

4 e-

5 e-

10 e-

Explanation

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

What is the coefficient for oxygen gas when the following equation is balanced?

$C_{5}H_{12} + O_{2} \rightarrow CO_{2} + H_{2}O$

Explanation

The balanced reaction for the combustion of pentane is:

$C_{5}H_{12} + 8O_{2} \rightarrow 5CO_{2} + 6H_{2}O$

When balanced, oxygen gas has a coefficient of eight.

To balance the equation, it is easiest to leave oxygen and hydrogen for last. This means we should start with carbon.

$C_{5}H_{12} + O_{2} \rightarrow CO_{2} + H_{2}O$

$C_{5}H_{12} + O_{2} \rightarrow 5CO_{2} + H_{2}O$

Now that carbon is balanced, we can look at hydrogen.

$C_{5}H_{12} + O_{2} \rightarrow 5CO_{2} + 6H_{2}O$

Finally, we can balance the oxygen.

$C_{5}H_{12} + 8O_{2} \rightarrow 5CO_{2} + 6H_{2}O$

The final reaction uses five carbon atoms, twelve hydrogen atoms, and sixteen oxygen atoms per side.

The following ReDox reaction takes place in acidic solution:

Fe2+ + Cr2O72– → Fe3+ + Cr3+

What is the sum of coefficients in this redox reaction?

Explanation

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–

What is the coefficient for oxygen gas when the following equation is balanced?

$C_{5}H_{12} + O_{2} \rightarrow CO_{2} + H_{2}O$

Explanation

The balanced reaction for the combustion of pentane is:

$C_{5}H_{12} + 8O_{2} \rightarrow 5CO_{2} + 6H_{2}O$

When balanced, oxygen gas has a coefficient of eight.

To balance the equation, it is easiest to leave oxygen and hydrogen for last. This means we should start with carbon.

$C_{5}H_{12} + O_{2} \rightarrow CO_{2} + H_{2}O$

$C_{5}H_{12} + O_{2} \rightarrow 5CO_{2} + H_{2}O$

Now that carbon is balanced, we can look at hydrogen.

$C_{5}H_{12} + O_{2} \rightarrow 5CO_{2} + 6H_{2}O$

Finally, we can balance the oxygen.

$C_{5}H_{12} + 8O_{2} \rightarrow 5CO_{2} + 6H_{2}O$

The final reaction uses five carbon atoms, twelve hydrogen atoms, and sixteen oxygen atoms per side.

How many electrons are involved in the following reaction?

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

5 e-

4 e-

3 e-

2 e-

1 e-

Explanation

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

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

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

Explanation

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:

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

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)

Explanation

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:

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

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

$K_3As$

$K_2As$

$KAs$

$KAs_3$

$KAs_2$

Explanation

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 $K_3As$ .

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

$K_3As$

$K_2As$

$KAs$

$KAs_3$

$KAs_2$

Explanation

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 $K_3As$ .

What is the balanced chemical equation for the combustion of butane $(C_4H_{10})$ ?

$2C_4H_{10} + 13O_2 \rightarrow 8CO_2 + 10H_2O$

$C_4H_{10} + 13O_2 \rightarrow 4CO_2 + 5H_2O$

$C_4H_{10} \rightarrow 4C + 5H_2$

$C_4H_{10} + 4O_2 \rightarrow 4CO_2 + 5H_2$

$C_4H_{10} + 4O_2 \rightarrow 4CO_2 + 8H_2$

Explanation

Combustion is the chemical reaction of a hydrocarbon with molecular oxygen, and it always produces carbon dioxide and water. Knowing the reactants and products, the unbalanced equation must be:

$C_4H_{10} + O_2 \rightarrow CO_2 + H_2O$

We start by balancing the hydrogens. Since there are 10 on the left and only 2 on the right, we put a coefficient of 5 on water.

$C_4H_{10} + O_2 \rightarrow CO_2 + 5H_2O$

Similarly, we balance carbons by putting a 4 on the carbon dioxide.

$C_4H_{10} + O_2 \rightarrow 4CO_2 + 5H_2O$

To find the number of oxygens on the right, we multiply the 4 coefficient by the 2 subscript on O (which gets us 8 oxygens) and then add the 5 oxygens from the 5 water molecules to get a total of 13. The needed coefficient for on the left would then have to be 13/2.