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Chapter 9: Problem 170

Calcium reacts with nitrogen gas to form calcium nitride. If \(33.8 \mathrm{~g}\) of calcium react with \(20.4 \mathrm{~g}\) of nitrogen gas, (a) Which reactant is the limiting reagent? (b) If the reaction has a \(72.4 \%\) yield, how many grams of calcium nitride are formed?

Short Answer

Expert verified
(a) The limiting reactant is calcium. (b) If the reaction has a 72.4% yield, 30.2 grams of calcium nitride are formed.

Step by step solution

01

Write the balanced chemical equation

The first step is to write down the balanced chemical equation for the reaction between calcium and nitrogen gas: \[ 3Ca + N_2 \rightarrow Ca_3N_2 \]
02

Calculate the moles of reactants

Next, convert the given masses of reactants into moles using the molar mass of the elements (Ca = 40.08 g/mol, N = 14.01 g/mol): \[\text{moles of calcium} = \frac{33.8\,\text{g}}{40.08\,\text{g/mol}} = 0.843\,\text{mol}\] \[\text{moles of nitrogen} = \frac{20.4\,\text{g}}{28.02\,\text{g/mol}} = 0.728\,\text{mol}\]
03

Determine the limiting reactant

Now, determine the limiting reactant by comparing the ratios of the moles of reactants to the coefficients in the balanced chemical equation: \[\frac{\text{moles of calcium}}{3} = \frac{0.843}{3} = 0.281\] \[\frac{\text{moles of nitrogen}}{1} = \frac{0.728}{1} = 0.728\] The ratio for calcium (0.281) is less than the ratio for nitrogen (0.728), so calcium is the limiting reactant. (a) The limiting reactant is calcium.
04

Calculate the theoretical yield

Determine the theoretical yield by finding the moles of calcium nitride that can be formed from the limiting reactant (calcium): \[\text{moles of calcium nitride} = \frac{0.843\,\text{mol Ca}}{3\,\text{mol Ca}} \times 1\,\text{mol}\,\text{Ca}_3\text{N}_2 = 0.281\,\text{mol}\] Next, convert the moles of calcium nitride into grams: \[\text{theoretical yield} = 0.281\,\text{mol} \times 148.25\,\text{g/mol} = 41.7\,\text{g}\]
05

Calculate the actual yield

Now, calculate the actual yield using the given percentage yield of 72.4%: \[ \text{actual yield} = \text{theoretical yield} \times \text{percentage yield} \] \[ \text{actual yield} = 41.7\,\text{g} \times 0.724 = 30.2\,\text{g} \] (b) If the reaction has a 72.4% yield, 30.2 grams of calcium nitride are formed.

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