How many grams of $C a O$ are required to neutralize $852$ gm of $P_{4} O_{10}$ ?

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Solution

$6 C a O + P_{4} O_{10} \to 2 (C a)_{3}$ $(P O_{4})_{2}$

852 gm $P_{4} O_{10} = 3$ moles $P_{4} O_{10}$

Since, 1 mole $o f P_{4} O_{10}$ neutralizes 6 moles of $C a O$ .

Therefore, 3 moles of $P_{4} O_{10}$ will neutralize $6 \times 3 = 18$ moles of $C a O$ .

Weight of $C a O$ required $= 18 \times 56 = 1008$ gm

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