Question

Mass of calcium oxide required when it completely reacts with $852$ g of $P_4{O}_{10}$ is:

• A. $100$ g
• B. $1008$ g
• C. $108$ g
• D. $1050$ g

Answer 1

The correct option is B $1008$ g

The reaction is as follows :

$6CaO+P_4{O}_{10}\to 2{Ca}_3{\left(P{O}_4\right)}_2$

$\because$ $1$ mole of $P_4{O}_{10}$ requires $6$ moles $CaO$.

$\therefore \frac{852}{284}$ moles of $P_4{O}_{10}$ require $\frac{6\times 852}{284}$ moles $CaO$ $=\frac{6\times 852}{284}\times 56$ g $CaO$.

Therefore, mass of $CaO$ required $=1008$ g

Hence, the correct option is $B$