Example 4.12: Identifying the Limiting Reactant
Identifying the Limiting Reactant
Silicon nitride is a very hard, high-temperature-resistant ceramic used as a component of turbine blades in jet engines. It is prepared according to the following equation:
\(\ce{3Si(s)}\)\(\ce{ + }\)\(\ce{2N2(g)}\)\(\ce{->}\)\(\ce{Si3N4(s)}\)\(\ce{ }\)
Which is the limiting reactant when
2.00 g of Si and
1.50 g of
\(\ce{N2}\) react?
Solution
Compute the provided chemical amounts of reactants, and then compare these amounts to the balanced equation to identify the limiting reactant.
\(m_{\mathrm{\ce{Si}}}\) \(= 2.00\ \mathrm{g}\)
\(m_{\mathrm{\ce{N2}}}\) \(= 1.50\ \mathrm{g}\)
\(n_{\mathrm{\ce{Si}}}\) \(= \dfrac{m_{\mathrm{\ce{Si}}}}{M_{\mathrm{\ce{Si}}}}\)
\(\ \ \ =\dfrac{2.00\ \mathrm{g}}{28.085\ \frac{\mathrm{g}}{\mathrm{mol}}}\)
\(\ \ \ =0.0712\ \mathrm{mol}\)
\(n_{\mathrm{\ce{N2}}}\) \(= \dfrac{m_{\mathrm{\ce{N2}}}}{M_{\mathrm{\ce{N2}}}}\)
\(\ \ \ =\dfrac{1.50\ \mathrm{g}}{28.014\ \frac{\mathrm{g}}{\mathrm{mol}}}\)
\(\ \ \ =0.0535\ \mathrm{mol}\)
\(n_{\mathrm{\ce{->}}}\) \(= \mathrm{min}(\dfrac{n_{\mathrm{\ce{Si}}}}{3}, \dfrac{n_{\mathrm{\ce{N2}}}}{2})\)
\(\ \ \ =\mathrm{min}(\dfrac{0.0712\ \mathrm{mol}}{3}, \dfrac{0.0535\ \mathrm{mol}}{2})\)
\(\ \ \ =\mathrm{min}(0.02374\ \mathrm{mol}, 0.02677\ \mathrm{mol})\)
\(\ \ \ =0.0237\ \mathrm{mol}\)
Comparing the ratios of chemical amount to stoichiometric coefficient shows that Si is limiting (smaller number, runs out while
\(\ce{N2}\) is left over).
To check our results, we calculate how much is left over.
\(n_{\mathrm{\ce{Si},leftover}}\) \(= n_{\mathrm{\ce{Si}}} - n_{\mathrm{\ce{->}}} \cdot 3\)
\(\ \ \ =0.0712\ \mathrm{mol} - 0.0237\ \mathrm{mol} \cdot 3\)
\(\ \ \ =0.0712\ \mathrm{mol} - 0.07121\ \mathrm{mol}\)
\(\ \ \ =0.0000\)
\(n_{\mathrm{\ce{N2},leftover}}\) \(= n_{\mathrm{\ce{N2}}} - n_{\mathrm{\ce{->}}} \cdot 2\)
\(\ \ \ =0.0535\ \mathrm{mol} - 0.0237\ \mathrm{mol} \cdot 2\)
\(\ \ \ =0.0535\ \mathrm{mol} - 0.04747\ \mathrm{mol}\)
\(\ \ \ =6.1\times 10^{-3}\ \mathrm{mol}\)
These chemical amounts should either be zero (for the limiting reactant) or larger than zero (for the reactants in excess). If we obtain any negative chemical amounts, we know there was a mistake in the setup or in the calculation.