Question

How many grams of $CaO$ are required to react with $852 \,g$ of $P_{4}O_{10}$?

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

Answer 1

Answer: (B)

$1008\,g$

Answer 2

$6CaO+P_{4}O_{10}\to2Ca_{3}\left(PO_{4}\right)_{2}$ $1$ mole of $P_{4}O_{10} =$ molar mass of $P_{4}O_{10} = 284 \,g$ $852 \,g$ 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 $CaO = 18\times 56 = 1008 g$