JEE MAIN - Chemistry (2022 - 28th July Morning Shift - No. 14)
In the given reaction,
$$X+Y+3 Z \leftrightarrows X YZ_{3}$$
if one mole of each of $$X$$ and $$Y$$ with $$0.05 \mathrm{~mol}$$ of $$Z$$ gives compound $$X Y Z_{3}$$. (Given : Atomic masses of $$X, Y$$ and $$Z$$ are 10, 20 and 30 amu, respectively.) The yield of $$X YZ_{3}$$ is _____________ g. (Nearest integer)
Answer
2
Explanation
$\underset{n_{\text {moles }}=1}{X}+\underset{1}{Y}+\underset{0.05}{3 Z} \rightleftharpoons X Y Z_{3}$
Limiting reagent is $Z=\frac{0.05}{3}=.016$
3 moles of $Z \rightarrow 1$ mole of $X Y Z_{3}$
$0.05$ mole of $Z \rightarrow \frac{1}{3} \times 0.05$ mole of $X Y Z_{3}$
M.wt. of $\mathrm{XYZ}_{3}=10+20+90$
$=120\, \mathrm{amu}$
Wt. of $X Y Z_{3}=\frac{.05}{3} \times 120$
$=2 \mathrm{~g}$
Limiting reagent is $Z=\frac{0.05}{3}=.016$
3 moles of $Z \rightarrow 1$ mole of $X Y Z_{3}$
$0.05$ mole of $Z \rightarrow \frac{1}{3} \times 0.05$ mole of $X Y Z_{3}$
M.wt. of $\mathrm{XYZ}_{3}=10+20+90$
$=120\, \mathrm{amu}$
Wt. of $X Y Z_{3}=\frac{.05}{3} \times 120$
$=2 \mathrm{~g}$
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