In the reaction $\mathrm{N}_{2} \mathrm{H}_{4}+3 \mathrm{O}_{2} ightarrow 2 \mathrm{NO}_{2}+2 \mathrm{H}_{2}…

In the reaction $\mathrm{N}_{2} \mathrm{H}_{4}+3 \mathrm{O}_{2} ightarrow 2 \mathrm{NO}_{2}+2 \mathrm{H}_{2} \mathrm{O}$, the mass of $\mathrm{O}_{2}$ required to combine with $745 \mathrm{~g}$ of $\mathrm{N}_{2} \mathrm{H}_{4}$ will be
  1. $2120 \mathrm{~g}$
  2. $2235 \mathrm{~g}$
  3. $2436 \mathrm{~g}$
  4. $2510 \mathrm{~g}$

Solution

We have
$\begin{array}{l}\mathrm{N}_{2} \mathrm{H}_{4}+3 \mathrm{O}_{2} ightarrow 2 \mathrm{NO}_{2}+2 \mathrm{H}_{2} \mathrm{O} \\ 32 \mathrm{~g} \quad\quad 96 \mathrm{~g}\end{array}$
Mass of $\mathrm{O}_{2}$ required $=\left(\frac{96 \mathrm{~g}}{32 \mathrm{~g}}ight)(745 \mathrm{~g})=2235 \mathrm{~g}$ /

Asked in: JEE-TOPICTESTS-CHEMISTRY

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