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 -4150 kJ/mol,how many grams of it must be used to heat 1.50 L of water from 20.0°C to 90.0°C? Assume all the energy is used to heat the water.

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First, we need to determine the amount of energy required to heat 1.50 L of water from 20.0°C to 90.0°C: q = (1.50 L)(1000 g/L)(4.184 J/g°C)(90.0°C - 20.0°C) q = 376,680 J Next, we need to determine the amount of C₆ required to produce this amount of energy: mol C₆ = (376,680 J) / (-4150 kJ/mol) mol C₆ = -90.8 mol Since we know that white gas is 100% C₆, the mass of white gas required can be determined using its molar mass: mass white gas = (-90.8 mol)(114 g/mol) mass white gas = 10,355 g = 10.36 g Therefore, the answer is C) 9.11 g (rounded to two significant figures).

Lightweight Camping Stoves Typically Use a Mixture of C5 and C6

Lightweight Camping Stoves Typically Use a Mixture of C₅ and C₆