A gas (Molar mass = 280 g mol⁻¹) was burnt in excess O₂ in a constant volume calorimeter and during combustion the temperature of calorimeter increased from 298.0 K to 298.45 K. If the heat capacity of calorimeter is 2.5 kJ K⁻¹ and enthalpy of combustion of gas is 9 kJ mol⁻¹ then amount of gas burnt is _____________ g. (Nearest Integer)

Solution

Chemistry ma’am · 3 steps

The calorimeter tells you the heat released; divide by heat per mole to get moles, then turn moles into grams.

1. Heat released. The heat given out by the burning gas is absorbed by the calorimeter, so q = C Δ T. Here Δ T = 298.45 - 298.0 = 0.45 K.

   q = 2.5 kJ K⁻¹ × 0.45 K = 1.125 kJ

2. Moles burnt. Each mole of gas releases 9 kJ, and the question takes this as the heat per mole. So moles burnt = 1.125/9 = 0.125 mol.
3. So. Mass = 0.125 × 280 = 35 g, so the answer is 35.

The common slip: 2240. Dividing 9 by 1.125 instead of 1.125 by 9 gives 8 mol and 2240 g. The total heat goes on top, the heat per mole goes below.

The idea to keep. In calorimetry, heat absorbed by the calorimeter equals heat released by the reaction. Heat divided by heat per mole gives moles, and moles times molar mass gives grams.

Answer from the NTA final answer key. Solution by PrepWiser.