Calculate the amount of $\mathrm{NO}_2$ required for producing $4 \mathrm{~g}$ moles of $\mathrm{HNO}_3$ as…
Calculate the amount of $\mathrm{NO}_2$ required for producing $4 \mathrm{~g}$ moles of $\mathrm{HNO}_3$ as per the chemical reaction, $3 \mathrm{NO}_2+\mathrm{H}_2 \mathrm{O} \longrightarrow 2 \mathrm{HNO}_3+\mathrm{NO}$. Given, the gram molecular weights of di-nitrogen and di-oxygen gases are $28 \mathrm{~g}$ and $32 \mathrm{~g}$ respectively.
$276 \mathrm{~g}$
$274 \mathrm{~g}$
$2 \mathrm{~g}$
$275 \mathrm{~g}$
Solution
$4 \mathrm{~g}$ mole $\mathrm{HNO}_3=4 \times 63=252 \mathrm{~g} \mathrm{HNO}_3$
$
\underset{(3 \times 46=138 \mathrm{~g})}{3 \mathrm{NO}_2+\mathrm{H}_2 \mathrm{O} \longrightarrow} \underset{(2 \times 63=126 \mathrm{~g})}{2} \mathrm{HNO}_3+\mathrm{NO}
$
According to the balance chemical reaction $126 \mathrm{~g} \mathrm{HNO}_3$ produced by $138 \mathrm{~g} \mathrm{NO}_2$ $\therefore 252 \mathrm{~g} \mathrm{HNO}_3$ will be produced by
$
\frac{126}{138} \times 252=276 \mathrm{~g} \mathrm{NO}_2
$
Hence, the correct option is (1)