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