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Q.
How many grams of $CaO$ are required to react with $852 \,g$ of $P_{4}O_{10}$?
Some Basic Concepts of Chemistry
Solution:
$6CaO+P_{4}O_{10}\to2Ca_{3}\left(PO_{4}\right)_{2}$
$1$ 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 $CaO = 18\times 56 = 1008 g$