Pls answer the question

Pls answer the question action Stoichiometry 50 g of CaC03 is allowed to react with 70 g of H3P04. Calculate : (ii) amount of unreacted reagent (i) amount of formed

Dear Student,

3 {CaCO}_{3} + 2 H_{3} {PO}_{4} \to {Ca}_{3} ({PO}_{4})_{2} + 3 {CO}_{2} + 3 H_{2} O 3 mol of {CaCO}_{3} reacts with 2 mol of H_{3} {PO}_{4} or, 3 \times 100 g of {CaCO}_{3} reacts with 2 \times 98 g of H_{3} {PO}_{4} So, 50 g of {CaCO}_{3} will react with \frac{2 \times 98}{3 \times 100} \times 50 = 32.66 g of H_{3} {PO}_{4} Hence, limiting reagent is {CaCO}_{3} (i) 3 \times 100 g of {CaCO}_{3} gives 310 g of {Ca}_{3} ({PO}_{4})_{2} So, 50 g of {CaCO}_{3} will give \frac{310}{300} \times 50 = 51.66 g of {Ca}_{3} ({PO}_{4})_{2} (ii) Since {CaCO}_{3} is limiting reagent so it will be completely consumed Amount of H_{3} {PO}_{4} left = 70 - 32.66 = 37.34 g