Question

How many grams of cao are required to react with 852g of p4o10?

Answer 1

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

The balanced 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.

But 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 = 18 x 56 = 1008 grams

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

hope it helps!!

Answer 2

Required Equation is

6CaO + P4O10 ➡️2Ca3(PO4)2

Here 6 mole of CaO requires of 1 mole of P4O10 for neutralisation

Molar mass of CaO=56 g

Molar mass of P4O10=284 g

i.e;6*56=336 g of need of CaO need 284 g of P4O10 for neutralisation

Then how many grams of CaO is neutralised by 852 g of P4O10

x=(336*852)/284=1008 g