Calculate the amount of calcium oxide (in grams) required when it reacts with 852 g of P4O10.

Question

Calculate the amount of calcium oxide (in grams) required when it reacts with 852 g of P4O10. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

6 mathrmCaO+ mathrmP4 mathrmO10 longrightarrow 2 mathrmCa3( mathrmPO4)2 1 Mole P4O10 = 284 g/mole Moles of P4O10 = (8 5 2/2 8 4) = 3 moles Moles of CaO = 18 moles Mass of CaO = 18 x 56 = 1008 g

Q. Calculate the amount of calcium oxide (in grams) required when it reacts with 852 g of P₄O₁₀.

$6 CaO + P_{4} O_{10} ⟶ 2 Ca_{3} (PO_{4})_{2}$

1 Mole P₄O₁₀ = 284 g/mole

Moles of P₄O₁₀ = $\frac{852}{284}$ = 3 moles

Moles of CaO = 18 moles

Mass of CaO = 18 x 56 = 1008 g

Q. Calculate the amount of calcium oxide (in grams) required when it reacts with 852 g of P4O10.

6CaO+P4​O10​⟶2Ca3​(PO4​)2​ 1 Mole P4O10 = 284 g/mole Moles of P4O10 = 284852​ = 3 moles Moles of CaO = 18 moles Mass of CaO = 18 x 56 = 1008 g

Q. Calculate the amount of calcium oxide (in grams) required when it reacts with 852 g of P₄O₁₀.

$6 CaO + P_{4} O_{10} ⟶ 2 Ca_{3} (PO_{4})_{2}$

1 Mole P₄O₁₀ = 284 g/mole

Moles of P₄O₁₀ = $\frac{852}{284}$ = 3 moles

Moles of CaO = 18 moles

Mass of CaO = 18 x 56 = 1008 g