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Calcium carbide (CaC2) is manufactured by combining lime (CaO) and carbon at high temperatures, as shown in the equation. Approximately how many grams of CaC2 can be formed from 84.0 g of CaO and 48.0 g of C?

CaO(s) + 3C(s) CaC2 + CO(g)

Calcium carbide (CaC2) is manufactured by combining lime (CaO) and carbon at high-example-1

1 Answer

1 vote

Answer:

A) 85.4 Grams

Step-by-step explanation:

To solve this problem, we need to determine which reactant is limiting and which reactant is in excess. The reactant that is limiting is the one that is completely consumed in the reaction, while the reactant that is in excess is the one that is left over after the limiting reactant is consumed.

To find out which reactant is limiting, we can use stoichiometry to calculate how much CaC2 can be formed from each reactant. We will then compare these amounts to determine which reactant produces the least amount of CaC2.

First, we need to convert the given masses of CaO and C to moles:

⇒ 84.0 g CaO × (1 mol CaO/56.08 g CaO) = 1.50 mol CaO

⇒ 48.0 g C × (1 mol C/12.01 g C) = 4.00 mol C

Next, we can use the balanced chemical equation to calculate the theoretical yield of CaC2 from each reactant:

⇒ From CaO: 1.50 mol CaO × (1 mol CaC2/1 mol CaO) = 1.50 mol CaC2

⇒ From C: 4.00 mol C × (1 mol CaC2/3 mol C) = 1.33 mol CaC2

Based on these calculations, we can see that the amount of CaC2 that can be formed from CaO is greater than the amount that can be formed from C. Therefore, C is the limiting reactant and CaO is in excess.

To calculate the actual yield of CaC2, we can use the amount of limiting reactant (C) and the stoichiometry of the balanced chemical equation:

4.00 mol C × (1 mol CaC2/3 mol C) × (64.10 g CaC2/1 mol CaC2) = 85.3 g CaC2

Therefore, approximately 85.3 grams of CaC2 can be formed from 84.0 g of CaO and 48.0 g of C.

Hence, the answer would be A. 85.4 g

User Oleksandr Oliynyk
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