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50 g of CaCO3 is allowed to react with 68.6 g of H3PO4 according to the reaction: 3CaCO3 + 2H3PO4 -> Ca3(PO4)2 + 3H2O + 3CO2 Which of the following statements are correct?

1. 51.67 g of salt is formed
2. Amount of unreacted reagent = 35.93 g
3. nCO2 formed = 0.5 moles
4. nH2O formed = 0.7 moles

Correct answer: nCO2 formed = 0.5 moles

Solution

n(CaCO3) = 50/100 = 0.5 mol; n(H3PO4) = 68.6/98 = 0.7 mol. Required ratio 3:2; for 0.5 mol CaCO3 need 0.333 mol H3PO4, but we have 0.7 mol. For 0.7 mol H3PO4 need 1.05 mol CaCO3, but we only have 0.5. So CaCO3 is limiting. From 0.5 mol CaCO3: n(CO2) = 0.5 mol (1:1 ratio), n(H2O) = 0.5 mol, n(Ca3(PO4)2) = 0.5/3 mol = 0.1667 mol => 0.1667*310 = 51.67 g salt. Excess H3PO4 = 0.7 - (2/3)*0.5 = 0.7 - 0.333 = 0.367 mol => 0.367*98 = 35.93 g. So options A (51.67 g salt), B (35.93 g unreacted), and C (0.5 mol CO2) are all correct; D is wrong (H2O = 0.5 mol, not 0.7). The single-answer form given in the original suggests option C is the 'correct' one highlighted, but A and B are also correct.

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