How can I identify the limiting reactant when 43.25 g of cac2 reacts with 33.71 g of water to produce ca (oh) to and c2h2?

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How can I identify the limiting reactant when 43.25 g of cac2 reacts with 33.71 g of water to produce ca (oh) to and c2h2?

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Answer 1 · 2014-12-15T16:44:05

$CaC_{2}$ $"CaC"""_2$ is the limiting reagent.

Explanation:

The limiting reagent is $C a C_{2}$ $CaC_2$ .

First, start with the balanced chemical equation

$C a C_{2} + 2 H_{2} O \to C_{2} H_{2} + C a {(O H)}_{2}$ $CaC_2 + 2H_2O -> C_2H_2 + Ca(OH)_2$

Notice that we have a $1 : 2$ $1:2$ mole ratio between $C a C_{2}$ $CaC_2$ and $H_{2} O$ $H_2O$ ; that is, every mole of the former used requires 2 moles of the latter.

Since the quantities of both reactants are given, and knowing that their molar masses are $62.0 \frac{g}{m o l}$ $62.0 g/(mol)$ (for $C a C_{2}$ $CaC_2$ ) and $18.0 \frac{g}{m o l}$ $18.0 g/(mol)$ (for $H_{2} O$ $H_2O$ ), we can determine the number of moles from

$n_{H_{2} O} = \frac{m_{H_{2} O}}{m o l a r m a s s_{H_{2} O}} = \frac{33.71 g}{18 \frac{g}{m o l}} = 1.87$ $n_(H_2O) = m_(H_2O)/(molarmass_(H_2O)) = (33.71 g)/(18 g/(mol)) = 1.87$ moles

$n_{C a C_{2}} = \frac{m_{C a C_{2}}}{m o l a r m a s s_{C a C_{2}}} = \frac{43.25 g}{64.0 \frac{g}{m o l}} = 0.68$ $n_(CaC_2) = m_(CaC_2)/(molarmass_(CaC_2)) = (43.25 g)/(64.0 g/(mol)) = 0.68$ moles

Notice that the number of moles of $C a C_{2}$ $CaC_2$ determined would require $2 \cdot 0.68 = 1.36$ $2 * 0.68 = 1.36$ moles of $H_{2} O$ $H_2O$ , less than what we have in the reaction. Therefore, $H_{2} O$ $H_2O$ is in excess, which means that $C a C_{2}$ $CaC_2$ is the limiting reagent.

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How can I identify the limiting reactant when 43.25 g of cac2 reacts with 33.71 g of water to produce ca (oh) to and c2h2?

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