Example 9.18: Volumes of Reacting Gases
Ammonia is an important fertilizer and industrial chemical. Suppose that a volume of
683 billion cubic feet of gaseous ammonia, measured at
25 °C and
1 atm, was manufactured. What volume of
\(\ce{H2(g)}\), measured under the same conditions, was required to prepare this amount of ammonia by reaction with
\(\ce{N2}\)?
\(\ce{N2(g)}\)\(\ce{ + }\)\(\ce{3H2(g)}\)\(\ce{->}\)\(\ce{2NH3(g)}\)\(\ce{ }\)
Solution
\(ν_{\mathrm{\ce{H2(g)}}}\) \(= 3\)
\(ν_{\mathrm{\ce{NH3(g)}}}\) \(= 2\)
\(\mathrm{ft}\) \(= 12 \cdot 2.54\ \mathrm{cm}\)
\(\ \ \ =30.5\ \mathrm{cm}\)
\(V_{\mathrm{\ce{NH3}}}\) \(= 683000000000 \cdot {\mathrm{ft}}^{3}\)
\(\ \ \ =683000000000 \cdot {(30.5\ \mathrm{cm})}^{3}\)
\(\ \ \ =683000000000 \cdot 2.832\times 10^{4}\ \mathrm{cm}^{3}\)
\(\ \ \ =1.93\times 10^{16}\ \mathrm{cm}^{3}\)
\(V_{\mathrm{\ce{H2}}}\) = ?
Because equal volumes of
\(\ce{H2}\) and
\(\ce{NH3}\) contain equal numbers of molecules and each three molecules of
\(\ce{H2}\) that react produce two molecules of
\(\ce{NH3}\), the ratio of the volumes of
\(\ce{H2}\) and
\(\ce{NH3}\) will be equal to 3:2. Two volumes of
\(\ce{NH3}\), in this case in units of billion ft^3, will be formed from three volumes of H2:
\(V_{\mathrm{\ce{H2}}}\) \(= V_{\mathrm{\ce{NH3}}} \cdot \dfrac{ν_{\mathrm{\ce{H2(g)}}}}{ν_{\mathrm{\ce{NH3(g)}}}}\)
\(\ \ \ =1.93\times 10^{16}\ \mathrm{cm}^{3} \cdot \dfrac{3}{2}\)
\(\ \ \ =1.93\times 10^{16}\ \mathrm{cm}^{3} \cdot \frac{3 }{ 2}\)
\(\ \ \ =2.90\times 10^{16}\ \mathrm{cm}^{3}\)
\(V_{\mathrm{\ce{H2}}}\) \(= \dfrac{V_{\mathrm{\ce{H2}}}}{{\mathrm{ft}}^{3}}\mathrm{\ \mathrm{ft^3}}\)
\(\ \ \ =\dfrac{2.90\times 10^{16}\ \mathrm{cm}^{3}}{{(30.5\ \mathrm{cm})}^{3}}\mathrm{\ \mathrm{ft^3}}\)
\(\ \ \ =\dfrac{2.90\times 10^{16}\ \mathrm{cm}^{3}}{2.832\times 10^{4}\ \mathrm{cm}^{3}}\mathrm{\ \mathrm{ft^3}}\)
\(\ \ \ =1.02\times 10^{12}\mathrm{\ \mathrm{ft^3}}\)
The manufacture of
683 billion ft^3 of
\(\ce{NH3}\) required
1020 billion ft^3 of
\(\ce{H2}\). (At
25 °C and
1 atm, this is the volume of a cube with an edge length of approximately
1.9 miles.)
Think about it: Why do we need a larger volume of
\(\ce{H2}\) than the volume of the produced
\(\ce{NH3}\)?
a) Because the molar mass of
\(\ce{NH3}\) is larger
b) Because
\(\ce{NH3}\) has more atoms than
\(\ce{H2}\)
c) Because
\(\ce{H2}\) is more reactive than
\(\ce{NH3}\)
d) Because they react in a 3:2 ratio