What mass of butane (M = 58.0 g mol⁻¹) must undergo complete combustion to raise the temperature of 100.0 g of water by 1.00 °C? Assume that there is no heat loss. - VCE - SSCE Chemistry - Question 8 - 2011 - Paper 1

The combustion of butane can be expressed by the balanced equation:

$C_4H_{10} + 13/2 O_2 \rightarrow 4 CO_2 + 5 H_2O$

This reaction releases heat. The standard enthalpy change of combustion (ΔH) for butane is approximately -2878 kJ mol⁻¹. Thus, the heat released per mole of butane is:

$\text{Heat per mole of butane} = 2878 imes 1000 ext{ J} = 2878000 ext{ J}$

To find the moles of butane needed to produce the heat required:

n = rac{q}{\Delta H} = \frac{418.0 ext{ J}}{2878000 ext{ J mol}^{-1}} = 1.45 \times 10^{-4} ext{ mol}

Using the molar mass of butane (M = 58.0 g mol⁻¹), we can find the mass (m) of butane needed:

$m = n imes M = 1.45 \times 10^{-4} imes 58.0 = 0.00841 ext{ g}$

Rounding to significant figures, the mass of butane required is:

$m \approx 8.44 ext{ g}$