Calculate the heat required to make 6.4 kg of CaC2 from CaO(s) and C(s) from the reaction CaO(.s.)+3C(.s.) arrow CaC2(.s.)CO(.g.) give that, Δ Hfo(.CaO.)=-151.6 kcal, Δ Hfo(.CaC2.)=-14.2 kcal, Δ Hfo(.CO.)=-26.4 kcal

Question

Calculate the heat required to make 6.4 kg of CaC2 from CaO(s) and C(s) from the reaction CaO(.s.)+3C(.s.) arrow CaC2(.s.)CO(.g.) give that, Δ Hfo(.CaO.)=-151.6 kcal, Δ Hfo(.CaC2.)=-14.2 kcal, Δ Hfo(.CO.)=-26.4 kcal (A) 5624 kcal (B) 86.24× 103 kcal (C) 1.11× 104 kcal (D) 1100 kcal

Tardigrade Admin · Accepted Answer

CaO(.s.)+3C(.s.) arrow CaC2(.s.)+CO(.g.) Δ H= Sigma H(.Product.)- Sigma H(.reactant.) =[.-14.2-(.-26.4.)].-[.-151.6.)+3× 0]. =111 kcal Molecular weight of CaC2=64 64gCaC2=111 kcal So 6.4 kg CaC2=(111 × 6.4 × 1 03/64)=1.11× 104 kcal.

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

5624 kcal

$86.24 \times 1 0^{3}$ kcal

$1.11 \times 1 0^{4}$ kcal

1100 kcal

Q. Calculate the heat required to make 6.4 kg of CaC2​ from CaO(s) and C(s) from the reaction CaO(s)+3C(s)→CaC2​(s)CO(g) give that, ΔHfo​(CaO)=−151.6 kcal, ΔHfo​(CaC2​)=−14.2 kcal, ΔHfo​(CO)=−26.4 kcal

5624 kcal

86.24×103 kcal

1.11×104 kcal

1100 kcal

CaO(s)+3C(s)→CaC2​(s)+CO(g) ΔH=ΣH(Product)−ΣH(reactant) =[−14.2−(−26.4)]−[−151.6)+3×0] =111 kcal Molecular weight of CaC2​=64 64gCaC2​=111 kcal So 6.4 kg CaC2​=64111×6.4×103​=1.11×104 kcal.

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

$86.24 \times 1 0^{3}$ kcal

$1.11 \times 1 0^{4}$ kcal