Question

Industrially, vanadium metal can be obtained by reacting vanadium (V) oxide with calcium at high temperatures as given below:
$5Ca+V_2{O}_5⟶5CaO+2V$
In one process, $1.54\times {10}^{3}g$ of $V_2{O}_5$ react with $1.96\times {10}^{3}g$ of $Ca$, where the theoretical yield of $V$ is $x$ and the per cent yield, if $803$ g of $V$, is $y$ $(Ca=40,V=51,O=16)$. The value of $x+y$ is (nearest integer value) :

Answer 1

The molar masses of $Ca$, $V_2{O}_5$ and $V$ are $40.1$ g/mol, $181.8$ g/mol and $50.9$ g/mol respectively.
The number of moles of $Ca$ are $\frac{1.96\times {10}^3}{40.1}=48.88$ moles.
The number of moles of $V_2{O}_5$ are $\frac{1.54\times {10}^3}{181.8}=8.47$ moles.
$1$ mole of $V_2{O}_5$ reacts with $5$ moles of $Ca$.
Hence, $8.47$ moles of $V_2{O}_5$ will react with $5\times 8.47=42.3$ moles of $Ca$.
However, $48.8$ moles of $Ca$ are present.
Thus, $V_2{O}_5$ is the limiting reagent.
$1$ mole of $V_2{O}_5$ forms $2$ moles of $V$.
Hence, $8.47$ moles of $V_2{O}_5$ will form $2\times 8.47=16.94$ moles of $V$.
Hence, the theoretical yield is $x=16.94$ moles $\times 50.9$ g/mol $=862.3$ g/mol.
The percentage yield is $y=\frac{803}{862.3}\times 100=93.2\%$.
Hence, $x+y=862.3+93.2=956$.