Question

50 grams of CaCO3 is allowed to react with 73.5 grams of H 3 PO 4 calculate the amount of Ca3 (PO 4)2 formed in moles

Answer 1

balanced equation- 3CaCo3+2H2Po4 gives Ca3(Po4)2+3Co2+3H2O  so

210g of CaCo3 and 196 H3Po4 gives 310g of Ca3(Po4)

According to problem,

50gof CaCo3 and 73.5 gH3Po4 gives x g of  Ca3(pO4) then ‘

x=(50+73.5)*310/210+196

i.e. equal to 75.66205g

Answer 2

Number of moles of CaCO₃ = given mass/molar mass = 50/100 = 0.5

Number of moles of H₃PO₄ = given mass/molar mass = 73.5/98 = 0.75

The balanced chemical equation is

3 CaCO₃ + 2 H₃PO₄ --------------> Ca₃(PO₄)₂ + 3 H₂CO₃

So, 3 moles of CaCO₃ reacts with 2 moles of H₃PO₄ to give 1 mole of Ca₃(PO₄)₂.

Now, 0.5 moles of CaCO₃ completely reacts with 0.33 moles of H₃PO₄ to give 0.16 moles of Ca₃(PO₄)₂.

Thus leaving (0.75-0.33) = 0.42 moles of H₃PO₄ unreacted.

Mass of the 0.16 moles of Ca₃(PO₄)₂ formed = 0.16 x molar mass

= 0.16 x 310 = 49.6 grams

∴ 49.6 grams or 0.16 moles of Ca₃(PO₄)₂ is formed.

Hope it helps!!