Lightweight camping stoves typically use a mixture of C₅ and C₆ hydrocarbons, which is called white gas. Assuming white gas is 100% C₆H₁₄ with an enthalpy of combustion of -4,150 kJ/mol, how many grams of it must be used to heat 1.50 L of water from 20°C to 90°C? Assume all the energy is used to heat the water.

Question

Quizplus · Accepted Answer

First, we need to calculate the amount of heat required to raise the temperature of 1.50 L of water from 20°C to 90°C: q = m*C*dT where q is the heat required, m is the mass of water, C is the specific heat capacity of water (4.184 J/g°C), and dT is the change in temperature. m = (1.50 L)(1000 mL/L)(1 g/mL) = 1500 g dT = 90°C - 20°C = 70°C q = (1500 g)(4.184 J/g°C)(70°C) = 439,440 J Next, we need to convert the energy required to heat the water from joules to kilojoules and then to the amount of C₅/white gas required: -439,440 J x (1 kJ/1000 J) x (1 mol C₆/ -4150 kJ) = -0.1059 mol C₆ Finally, we can use the molar mass of C₆ (86.18 g/mol) to calculate the mass of C₅/white gas required: -0.1059 mol x (86.18 g/mol) = -9.11 g Therefore, the answer is C, 9.11 g.

Lightweight Camping Stoves Typically Use a Mixture of C5 and C6

Lightweight Camping Stoves Typically Use a Mixture of C₅ and C₆