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In Ostwald's process for the manufacture of nitric acid, the first step involves the oxidation of ammonia gas by oxygen gas to give nitric oxide gas and steam. What is the maximum weight of nitric oxide that can be obtained starting only with \(10.0 \mathrm{~g}\) of ammonia and \(20.0\) g of oxygen?
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The balanced equation for the reaction is:
\(\begin{array}{lcc}4 \mathrm{NH}_3(\mathrm{~g}) & +5 \mathrm{O}_2(\mathrm{~g}) \longrightarrow & \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_2 \mathrm{O}(\mathrm{g}) \\ 4 \times 17 & 5 \times 32 & 4 \times 30 \\ =68 \mathrm{~g} & =160 \mathrm{~g} & =120 \mathrm{~g}\end{array}\)
Here, \(160 \mathrm{~g}\) of \(\mathrm{O}_2\) will react with \(\mathrm{NH}_3=68 \mathrm{~g}\)
\(20 \mathrm{~g}\) of \(\mathrm{O}_2\) will react with \(\mathrm{NH}_3=(68 / 160) \times 20=8 \cdot 5 \mathrm{~g}\)
Thus, \(\mathrm{O}_2\) is the limiting reagent, therefore, the calculations must be based upon the amount of \(\mathrm{O}_2\) taken and not on the amount of \(\mathrm{NH}_3\) taken.
From the equation,
\(160 \mathrm{~g}\) of \(\mathrm{O}_2\) produce \(\mathrm{NO}=120 \mathrm{~g}\)
\(20 \mathrm{~g}\) of \(\mathrm{O}_2\) will produce \(\mathrm{NO}=(120 / 160) \times 20=15 \mathrm{~g}\)
\(\begin{array}{lcc}4 \mathrm{NH}_3(\mathrm{~g}) & +5 \mathrm{O}_2(\mathrm{~g}) \longrightarrow & \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_2 \mathrm{O}(\mathrm{g}) \\ 4 \times 17 & 5 \times 32 & 4 \times 30 \\ =68 \mathrm{~g} & =160 \mathrm{~g} & =120 \mathrm{~g}\end{array}\)
Here, \(160 \mathrm{~g}\) of \(\mathrm{O}_2\) will react with \(\mathrm{NH}_3=68 \mathrm{~g}\)
\(20 \mathrm{~g}\) of \(\mathrm{O}_2\) will react with \(\mathrm{NH}_3=(68 / 160) \times 20=8 \cdot 5 \mathrm{~g}\)
Thus, \(\mathrm{O}_2\) is the limiting reagent, therefore, the calculations must be based upon the amount of \(\mathrm{O}_2\) taken and not on the amount of \(\mathrm{NH}_3\) taken.
From the equation,
\(160 \mathrm{~g}\) of \(\mathrm{O}_2\) produce \(\mathrm{NO}=120 \mathrm{~g}\)
\(20 \mathrm{~g}\) of \(\mathrm{O}_2\) will produce \(\mathrm{NO}=(120 / 160) \times 20=15 \mathrm{~g}\)
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