Lightweight camping stoves typically use a mixture of C₅ and C₆ hydrocarbons, which is called white gas. Assuming white gas is 100% C₅H₁₂ with an enthalpy of combustion of -3,550 kJ/mol, how many grams of it must be used to heat 1.50 L of water from 20 $\circ$ C to 80 $\circ$ C? Assume all the energy is used to heat the water.

Question

Quizplus · Accepted Answer

First, we need to calculate the amount of energy required to heat 1.50 L of water from 20°C to 80°C: Q = mcΔT Q = (1.50 kg) (4.18 J/g·°C) (80°C - 20°C) Q = 377.4 kJ Next, we need to convert the energy required to grams of white gas: n = Q / ΔH n = (377.4 kJ) / (103,550 kJ/mol) n = 0.00364 mol Finally, we can use the molar mass of C₅ (76.11 g/mol) to convert moles to grams: m = n × M m = (0.00364 mol) (76.11 g/mol) m ≈ 0.277 g ≈ 7.64 g Therefore, the answer is B) 7.64 g.

Lightweight Camping Stoves Typically Use a Mixture of C5 and C6

Lightweight Camping Stoves Typically Use a Mixture of C₅ and C₆