## Chemistry: The Molecular Science (5th Edition)

Heat transfer = $7.251 \times 10^6 kJ$
40.08* 1 + 12.01* 2 = 64.10g/mol 1000 kg = $1000 * 10^3 g$ = $1.000 \times 10^6g$ 1. Calculate the amount in moles: $1.000\times 10^{6}g \times \frac{1 mol (CaC_2)}{ 64.10} = 1.560\times 10^{4} mol$ 2. Divide the amount by the coefficient of $CaC_2$: $\frac{1.560\times 10^{4}}{ 1} = 1.560\times 10^{4} mol - reaction$ 3. Find the necessary heat: $1.560\times 10^{4}mol-reaction \times 464.8 kJ/mol-reaction = 7.251\times 10^{6}kJ$