Question

How many grams of $CaO$ are required to react with $852g$ of $P_4{O}_{10}$?

• A. $852g$
• B. $1008g$
• C. $85g$
• D. $7095g$

Answer 1

Answer: (B)

$6CaO+P_4{O}_{10}\to 2C{a}_3{\left(P{O}_4\right)}_2$
$1$ mole of $P_4{O}_{10}=$ molar mass of $P_4{O}_{10}=284g$
$852g$ of$P_4{O}_{10}=\frac{852}{284}=3$mol
$1$ mole of $P_4{O}_{10}$ reacts with $6$ moles of $CaO$
$3$ moles of $P_4{O}_{10}$ reacts with $18$ moles of $CaO$
Mass of $18$ moles of $CaO=18\times 56=1008g$