Example 4.10: Relating Masses of Reactants and Products
Relating Masses of Reactants and Products
What mass of sodium hydroxide, \(\ce{NaOH}\), would be required to produce 16 g of the antacid milk of magnesia [magnesium hydroxide, \(\ce{Mg(OH)2]}\) by the following reaction?\(\ce{MgCl2(aq)}\)\(\ce{ + }\)\(\ce{2NaOH(aq)}\)\(\ce{->}\)\(\ce{Mg(OH)2(s)}\)\(\ce{ + }\)\(\ce{2NaCl(aq)}\)\(\ce{ }\)
Solution
\(m_{\mathrm{\ce{Mg(OH)2}}}\) \(= 16.\ \mathrm{g}\)The approach used previously in Example 7 and Example 8 is likewise used here; that is, we must derive an appropriate stoichiometric factor from the balanced chemical equation and use it to relate the amounts of the two substances of interest. In this case, however, masses (not molar amounts) are provided and requested, so additional steps of the sort learned in the previous chapter are required. The calculations required are outlined in this flowchart:
\(\dfrac{n_{\mathrm{\ce{NaOH}}}}{n_{\mathrm{\ce{Mg(OH)2}}}} = \dfrac{2}{1}\)
\(n_{\mathrm{\ce{Mg(OH)2}}}\) \(= \dfrac{m_{\mathrm{\ce{Mg(OH)2}}}}{M_{\mathrm{\ce{Mg(OH)2}}}}\)
\(\ \ \ =\dfrac{16.\ \mathrm{g}}{58.319\ \frac{\mathrm{g}}{\mathrm{mol}}}\)
\(\ \ \ =0.27\ \mathrm{mol}\)
\(n_{\mathrm{\ce{NaOH}}}\) \(= 2 \cdot n_{\mathrm{\ce{Mg(OH)2}}}\)
\(\ \ \ =2 \cdot 0.27\ \mathrm{mol}\)
\(\ \ \ =0.55\ \mathrm{mol}\)
\(m_{\mathrm{\ce{NaOH}}}\) \(= n_{\mathrm{\ce{NaOH}}} \cdot M_{\mathrm{\ce{NaOH}}}\)
\(\ \ \ =0.55\ \mathrm{mol} \cdot 39.997\ \frac{\mathrm{g}}{\mathrm{mol}}\)
\(\ \ \ =22.\ \mathrm{g}\)