MCQ
In the reaction $H_2 + O_2 \to H_2O$. If $6\,g$ of $H_2$ combines with $64\,g$ of $O_2$. Find mass of Excess reagent Left ? ............. $\mathrm{g}$
- A$32 $
- B$48 $
- ✓$16 $
- DNone
✓
Answer
Correct option: C.
$16 $
c
$\mathop {\mathop {2{{\text{H}}_2}}\limits_{6\,g} }\limits_{(3\,mol)} + \mathop {\mathop {{{\text{O}}_2}}\limits_{64\,g} }\limits_{(2\,mol)} \to 2{{\text{H}}_2}{\text{O}}$
$\mathop {\mathop {2{{\text{H}}_2}}\limits_{6\,g} }\limits_{(3\,mol)} + \mathop {\mathop {{{\text{O}}_2}}\limits_{64\,g} }\limits_{(2\,mol)} \to 2{{\text{H}}_2}{\text{O}}$
$[\mathrm{L} \cdot \mathrm{R.}]$
$2 \,\mathrm{mol}\, \mathrm{H}_{2}$ reacts with $\mathrm{O}_{2}=1\, \mathrm{mol}$
$\therefore $ $3\, \mathrm{mol}\, \mathrm{H}_{2}$ will react with $\mathrm{O}_{2}=\frac{1}{2} \times 3=1.5 \,\mathrm{mol}$
remaining amount of $\mathrm{O}_{2}=0.5\, \mathrm{mol}=16\, \mathrm{g}$
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