Question

consider the unbalanced equation for the combustion of butane

c4h10(g) + o2(g) → co2(g) + h2o(g)

part a

balance the equation

express your answer as a chemical equation including phases.

part b

determine how many moles of o2 are required to react completely with 7.6 mol of c4h10.

express your answer in moles to two significant figures.

Explanation:

Step1: Balance the chemical equation

For \(C_4H_{10}(g)+O_2(g)\to CO_2(g)+H_2O(g)\)

• Balance \(C\) atoms: \(C_4H_{10}(g)+O_2(g)\to 4CO_2(g)+H_2O(g)\)
• Balance \(H\) atoms: \(C_4H_{10}(g)+O_2(g)\to 4CO_2(g)+5H_2O(g)\)
• Balance \(O\) atoms: \(C_4H_{10}(g)+\frac{13}{2}O_2(g)\to 4CO_2(g)+5H_2O(g)\)

Multiply through by 2 to get whole - number coefficients: \(2C_4H_{10}(g)+13O_2(g)\to 8CO_2(g)+10H_2O(g)\)

Step2: Use stoichiometry for Part B

From the balanced equation \(2C_4H_{10}(g)+13O_2(g)\to 8CO_2(g)+10H_2O(g)\), the mole ratio of \(C_4H_{10}\) to \(O_2\) is \(n(C_4H_{10}):n(O_2)=2:13\)
If \(n(C_4H_{10}) = 7.6\space mol\), then \(n(O_2)=\frac{13}{2}\times n(C_4H_{10})\)
\(n(O_2)=\frac{13}{2}\times7.6\space mol = 49.4\space mol\approx49\space mol\) (to two significant figures)

Answer:

Part A: \(2C_4H_{10}(g)+13O_2(g)\to 8CO_2(g)+10H_2O(g)\)
Part B: \(49\space mol\)