Grade 10Physical Chemistry
Calculate the amount of calcium oxide required when it reacts with 852 g of P4=O10.
3 Answers
Deepak Patra
Sol. The balanced reaction is :
6CaO + P4O10 → 2Ca3 (PO4)2
Moles of P4O10 = 852/284 = 3
Moles of CaO required = 3 × 6 = 18
Mass of CaO required = 18 × 56
= 1008 g
samudra
reaction
here,
no. of moles of P4O10=852/284=3
thus,3 moles of CaO is also required
now,
3=M/(6*56)
therefore, 1008g of CaO is required
ankit singh
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. Therefore 1008 grams of CaO is required to react with 852 grams of P₄O₁₀.
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