Question

how many grams of CaO are required to neutralize 852gm of P4O10​

Answer 1

"Answer:

1008 g of CaO

Explanation:

Number of moles of P₄O₁₀ in 852 grams = 852/284 = 3 moles

The chemical reaction is;

6 CaO + P₄O₁₀ -----------> 2 Ca₃(PO₄)₂

6 moles of CaO reacts with 1 mole of P₄O₁₀ to give 2 moles of Ca₃(PO₄)₂.

There are 3 moles of P₄O₁₀ present in the sample.

So, 18 moles of CaO reacts with 3 moles of P₄O₁₀ to give 6 moles of Ca₃(PO₄)₂.

Mass of 18 moles of CaO will be, 18 x 56 = 1008 grams

Therefore 1008 grams of CaO is required to react with 852 grams of P₄O₁₀.

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