Answer:
Volume of \[{{H}_{2}}S{{O}_{4}}\] used
= 10 mL of \[1M\,\,{{H}_{2}}S{{O}_{4}}\]
Now,\[{{H}_{2}}S{{O}_{4}}+2N{{H}_{3}}\xrightarrow{{}}{{(N{{H}_{4}})}_{2}}S{{O}_{4}}\]
So, 1 mole
of \[{{H}_{2}}S{{O}_{4}}\] reacts with 2 moles of \[N{{H}_{3}}\]
i.e., 10 mL
of 1 M,\[S{{O}_{2}}\] = 20 mL of 1 M\[N{{H}_{3}}\]
Now, 1000 mL
of 1 M\[N{{H}_{3}}\] contains\[N=14g\]
So, 20 mL of
1 M \[N{{H}_{3}}\] contains \[N=\frac{14\times 20}{100}\]
Percentage
of nitrogen is = \[\frac{1.4\times N\times V}{W}\]
\[=\frac{14\times
20\times 100}{100\times 0.5}=56%\]
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