Question

on a 14 day wilderness expedition youll need to heat 3.0 kg of water to the boiling point each day. the air temperature will average 25 °c. you have available canisters of compressed butane (\\(\ce{c4h10}\\)) fuel, which youll burn to heat the water. each canister has 100. g of butane in it. what is the minimum number of fuel canisters you must bring? the standard heat of formation of butane at 25 °c is -125.7 kj/mol. youll probably find other helpful data in the aleks data resource. canisters:

Explanation:

Step1: Calculate total water mass

Total mass of water over 14 days:
$m_{\text{water}} = 3.0\ \text{kg/day} \times 14\ \text{days} = 42\ \text{kg} = 42000\ \text{g}$

Step2: Find heat to boil water

Use specific heat capacity of water $c = 4.184\ \text{J/g}^\circ\text{C}$, boiling point $100^\circ\text{C}$.
Heat needed:
$$q = m_{\text{water}} \times c \times \Delta T = 42000\ \text{g} \times 4.184\ \text{J/g}^\circ\text{C} \times (100-25)^\circ\text{C}$$
$$q = 42000 \times 4.184 \times 75 = 1.31772 \times 10^7\ \text{J} = 13177.2\ \text{kJ}$$

Step3: Write butane combustion reaction

Combustion reaction:
$\text{C}_4\text{H}_{10}(g) + \frac{13}{2}\text{O}_2(g)
ightarrow 4\text{CO}_2(g) + 5\text{H}_2\text{O}(l)$

Step4: Calculate $\Delta H^\circ_{\text{rxn}}$

Use standard heats of formation: $\Delta H^\circ_f(\text{C}_4\text{H}_{10})=-125.7\ \text{kJ/mol}$, $\Delta H^\circ_f(\text{CO}_2)=-393.5\ \text{kJ/mol}$, $\Delta H^\circ_f(\text{H}_2\text{O}(l))=-285.8\ \text{kJ/mol}$, $\Delta H^\circ_f(\text{O}_2)=0$.
$$\Delta H^\circ_{\text{rxn}} = [4\Delta H^\circ_f(\text{CO}_2) + 5\Delta H^\circ_f(\text{H}_2\text{O})] - [\Delta H^\circ_f(\text{C}_4\text{H}_{10}) + \frac{13}{2}\Delta H^\circ_f(\text{O}_2)]$$
$$\Delta H^\circ_{\text{rxn}} = [4(-393.5) + 5(-285.8)] - [-125.7 + 0]$$
$$\Delta H^\circ_{\text{rxn}} = [-1574 - 1429] + 125.7 = -2877.3\ \text{kJ/mol}$$
(negative sign means heat is released per mole of butane)

Step5: Molar mass of butane

Molar mass of $\text{C}_4\text{H}_{10}$:
$M = 4(12.01) + 10(1.008) = 58.12\ \text{g/mol}$

Step6: Heat per canister

Heat from 100g butane:
$$q_{\text{canister}} = \frac{100\ \text{g}}{58.12\ \text{g/mol}} \times 2877.3\ \text{kJ/mol} \approx 4950.6\ \text{kJ}$$

Step7: Number of canisters

Number of canisters (round up to whole number):

$$n = \frac{13177.2\ \text{kJ}}{4950.6\ \text{kJ/canister}} \approx 2.66 ightarrow 3$$

Answer:

3