Question

Calculate the heat required to make 6.4 kg of $CaC_{2}$ from CaO(s) and C(s) from the reaction $CaO\left(\right.s\left.\right)+3C\left(\right.s\left.\right) \rightarrow CaC_{2}\left(\right.s\left.\right)CO\left(\right.g\left.\right)$ give that, $\Delta H_{f}^{o}\left(\right.CaO\left.\right)=-151.6$ kcal, $\Delta H_{f}^{o}\left(\right.CaC_{2}\left.\right)=-14.2$ kcal, $\Delta H_{f}^{o}\left(\right.CO\left.\right)=-26.4$ kcal

• A. 5624 kcal
• B. $86.24\times 10^{3}$ kcal
• C. $1.11\times 10^{4}$ kcal
• D. 1100 kcal

Answer 1

Answer: (C)

$CaO\left(\right.s\left.\right)+3C\left(\right.s\left.\right) \rightarrow CaC_{2}\left(\right.s\left.\right)+CO\left(\right.g\left.\right)$

$\Delta H=\Sigma H\left(\right.Product\left.\right)-\Sigma H\left(\right.reactant\left.\right)$

$=\left[\right.-14.2-\left(\right.-26.4\left.\right)\left]\right.-\left[\right.-151.6\left.\right)+3\times 0\left]\right.$

$=111$ kcal

Molecular weight of $CaC_{2}=64$

$64gCaC_{2}=111$ kcal

So 6.4 kg $CaC_{2}=\frac{111 \times 6.4 \times 1 0^{3}}{64}=1.11\times 10^{4}$ kcal.