Question

44.8g of CaO is reacted with 28.4g of P4O10. Mass of calcium phosphate obtained in the reaction is
a. 85g
b. 75g
c. 73.2g
d. 62g​

Answer 1

Answer:

Mass of  $Ca_{2}$$(PO_{4}) _{2}$ is equal to 62 grams.

Explanation:

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

Molar mass of CaO =56g/mol

∴no.of moles of Ca0=$\frac{mass of CaO(in grams)}{Molar mass. of Cao(gram/mol)}$=$\frac{44.8}{56}$=0.8 moles

Molar mass of $P_{4}$$O_{10}$=284g/mol

∴no.of moles of $P_{4} O_{10}$=$\frac{mass of P_{4} 0_{10} (in grams)}{Molar mass. of P_{4} 0_{10} (gram/mol)}$=$\frac{28.4}{284}$=0.1 moles

from the reaction it is clear that the limiting reagent is $P_{4} O_{10}$(only 0.1 mole) available.

Hence,0.2 mole of  $Ca_{2}$$(PO_{4}) _{2}$ will be obtained as product,

molar mass of  $Ca_{2}$$(PO_{4}) _{2}$ is 310.

∴Mass of  $Ca_{2}$$(PO_{4}) _{2}$ obtained = (0.2)(310)=62 grams.