Question

When 0.103 g of Zn (s) is combined with enough HCl to make 50.0 mL of solution in a coffee cup calorimeter, all of the zinc reacts, raising the temperature of the solution from 22.5 degree C to 23.7 degree C. Find Delta H_rxn per mole of Zn. (Use 1.02 glmL for the density of the solution and 4.18 J/g degree C for the specific heat capacity.)
Zn (s) + 2 HCl (aq) rightarrow ZnCl_2 (aq) + H_2 (g)
Delta H_rxn/mol Zn = ____kJ/mol.

Answer 1

Answer:

$\Delta H_{rxn}=162940 J/mol$

Explanation:

To calculate the heat of reaction we need to determine:

1. The heat produced
2. The moles of Zn that reacted

Moles of zinc (Zn)

It says that the 0.103 g reacted completely, knowing that the Zn molecular weight is 65.4 g/mol:

$n_{Zn}=\frac{0.103g}{65.4 g/mol}=1.57*10^{-3}mol$

Heat released by the reaction

The solution goes from 22.5 °C to 23.7°C:

$Q=Cp*\rho *V*(T_f - T_i)$

$Q=4.18 \frac{J}{g* ^{\circ}C}*1.02 g/mL*50mL*(23.7-22.5)^{\circ}C$

$Q=255.8 J$

Heat of reaction

$\Delta H_{rxn}=\frac{-Q}{n_{Zn}}$

$\Delta H_{rxn}=\frac{-255.8 J}{1.57*10^-3}}=162940 J/mol$