Question

1. the following reaction represents the complete combustion of hexane, ( c_{6}h_{14} ). (watch your units!!!)

( c_{6}h_{14}(l)+\frac{19}{2}o_{2}(g)\to6co_{2}(g)+7h_{2}o(l)quad h = - 4163mathrm{kj} )
a. if 25.0kg of hexane is burned, how much heat will be released?
b. what mass of hexane is required to produce ( 1.00\times10^{5}mathrm{j} ) of heat?

Explanation:

Step1: Calculate the molar mass of hexane

The molar mass of \(C_6H_{14}\), \(M=(6\times12 + 14\times1)\text{ g/mol}=86\text{ g/mol}\)

Step2: Convert mass to moles for part a

Given \(m = 25.0\text{ kg}=25000\text{ g}\), the number of moles \(n=\frac{m}{M}=\frac{25000\text{ g}}{86\text{ g/mol}}\approx290.7\text{ mol}\)

Step3: Calculate heat released for part a

From the reaction \(C_6H_{14}(l)+\frac{19}{2}O_2(g)\to6CO_2(g) + 7H_2O(l)\), \(\Delta H=- 4163\text{ kJ}\) per \(1\text{ mol}\) of \(C_6H_{14}\)
Heat released \(q=n\times|\Delta H|=290.7\text{ mol}\times4163\text{ kJ/mol}\)
\(q = 290.7\times4163\times10^{3}\text{ J}\approx1.18\times10^{7}\text{ J}\)

Step4: Calculate moles for part b

Given \(q = 1.00\times10^{5}\text{ J}=100\text{ kJ}\)
Number of moles \(n=\frac{q}{|\Delta H|}=\frac{100\text{ kJ}}{4163\text{ kJ/mol}}\approx0.024\text{ mol}\)

Step5: Calculate mass for part b

Mass \(m=n\times M=0.024\text{ mol}\times86\text{ g/mol}=2.064\text{ g}\approx2.10\text{ kg}\) (after unit conversion)

Answer:

a. \( 1.18\times10^{7}\text{ J}\)
b. \( 2.10\text{ kg}\)