Question

Equal weight of ${}^{\prime }{X}^{\prime }$ (At. wt. = 36) and ${}^{\prime }{Y}^{\prime }$ (At. wt. =24) are reacted to form the compound $X_2{Y}_3$. Then:

• A. $X$ is the limiting reagent
• B. $Y$ is the llimiting reagent
• C. No reactant is left over and mass of $X_2{Y}_3$ formed is double the mass of ${}^{\prime }{X}^{\prime }$ taken
• D. None of these

Answer 1

The correct option is C No reactant is left over and mass of $X_2{Y}_3$ formed is double the mass of ${}^{\prime }{X}^{\prime }$ taken

Let ${}^{\prime }{w}^{\prime }$ is the weight of $X$ and $Y$ taken,
$\underset{\text{w gram}}{2X}+\underset{\text{w gram}}{3Y}\to \underset{0}{X_2{Y}_3}$
so,
Moles of $X=\frac{w}{36}andY=\frac{w}{24}$
To find Limiting Reagent :
$\frac{\text{given moles}}{\text{stoichiomteric coefficient}}$
$\to forX=\frac{w}{36}\times \frac{1}{2}=\frac{w}{72}forY=\frac{w}{24\times 3}=\frac{w}{72}$

So, no one is Limiting Reagent both are getting consumed completely.
$2X+3Y\to X_2{Y}_3$
moles of $X_2{Y}_3$ formed $=\frac{1}{2}\times \text{moles of X}=\frac{w}{2\times 36}$

$\text{so, weight of}{X}_2{Y}_3=moles\times molarmass=\frac{w}{2\times 36}(72+72)=2w$

So, weight of $X_2{Y}_3$ is double the weight of ${}^{\prime }{X}^{\prime }$ taken.