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100gm of Sample of caco3 reacts with 98grmsе оf H3PO4 -calculatea) no of grams of Ca 3 (PO4)2 that couldbe producedb) the no of grams of excess reagent thatwill remain unreacted |
| Answer» NFO : 100gm of Sample of CaCO3 reacts with 98 g оf H3PO4.To find : no of grams of Ca3(PO4)2 that could be produced the no of grams of EXCESS reagent that will REMAIN unreacted. solution : mass of CaCO3 = 100g molar mass of CaCO3 = 100g/mol so no of moles of CaCO3 = 100/100 = 1 similarly, mass of H3PO4 = 98g molar mass of H3PO4 = 68g/molso no of moles of H3PO4 = 1 reaction between CaCO3 and H3PO4 is..3CaCO3 + 2H3PO4 => Ca3(PO4)2 + 3H2O + 3CO2we see , three moles of CaCO3 react with 2 moles of H3PO4. so, 1 mole of CaCO3 will react with 2/3 mole of H3PO4. we see H3PO4 is in excess amount and hence CaCO3 is limiting reagent (because it is lesser in amount) so, CaCO3 will dominant to form Ca3(PO4)2 from reaction, it is clear that 3 moles of CaCO3 form 1 mole of Ca3(PO4)2 so, 1 mole of CaCO3 will form 1/3 mole of Ca3(PO4)2.so the mass of Ca3(PO4)2 = no of moles × molar mass = 1/3 mol × 310 g/mol [ ∵ molar mass of Ca3(PO4)2 is 310 g/mol ] = 103.33 g Therefore 103.33 g of Ca3(PO4)2 could be produced. we have seen that H3PO4 is in excess. so unreacted no of moles of H3PO4 = 1 - 2/3 = 1/3 mass of 1/3 moles of H3PO4 = 1/3 × molar mass of H3PO4 = 1/3 × 98 = 32.67g Therefore 32.67g of excess reagent (H3PO4) that will remained unreacted. | |