Question #f8d64

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Question #f8d64

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Answer 1 · 2014-10-25T01:24:10

You must first be sure the equation is balanced, which it is. Then you must determine the limiting reactant, as that will determine the theoretical yield of calcium carbonate.

$CaO (s)$ $"CaO"(s)$ + ${CO}_{2} (g)$ $"CO"_2(g)$ $\to$ $rarr$ ${CaCO}_{3} (s)$ $"CaCO"_3(s)$

The mole ratios for this equation are 1:1.

Determine the number of moles of each reactant. Moles are calculated by dividing the mass of each reactant by its molar mass. The molar mass of $CaO(s) = 56.077 g/mol$ $"CaO(s)" = "56.077 g/mol"$ . The molar mass of
${CO}_{2} (g)$ $"CO"_2(g)"$ = $44.099g/mol$ $"44.099g/mol"$

Moles of CaO.

$14.4g CaO(s)$ $"14.4g CaO(s)"$ x $\frac{1 mol CaO(s)}{56.077 g/mol}$ $"1 mol CaO(s)"/"56.077 g/mol"$ = $0.257 mol CaO(s)$ $"0.257 mol CaO(s)"$

Moles of ${CO}_{2} (g)$ $"CO"_2(g)"$ .

${13.8g CO}_{2} (g)$ $"13.8g CO"_2(g)$ x $\frac{1 mol}{44.099g/mol}$ $"1 mol"/"44.099g/mol"$ = ${0.313 mol CO}_{2}$ $"0.313 mol CO"_2$

Since there is less $CaO(s)$ $"CaO(s)"$ , it is the limiting reactant.

Since the mole ratio of $CaO(s)$ $"CaO(s)"$ and the product ${CaCO}_{3} (s)$ $"CaCO"_3(s)$ is 1:1, the reaction can produce no more than ${0.257 mol CaCO}_{3} (s)$ $"0.257 mol CaCO"_3(s)$

To determine the theoretical yield of ${CaCO}_{3} (s)$ $"CaCO"_3(s)$ in grams, multiply the number of moles of ${CaCO}_{3} (s)$ $"CaCO"_3(s)$ times its molar mass.

${0.257 mol CaCO}_{3} (s)$ $"0.257 mol CaCO"_3(s)$ x $\frac{100.11g}{1 mol}$ $"100.11g"/"1 mol"$ = ${25.72g CaCO}_{3} (s)$ $"25.72g CaCO"_3(s)$

The theoretical yield of ${CaCO}_{3} (s)$ $"CaCO"_3(s)$ in this experiment is $25.72g$ $"25.72g"$

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