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Calcium carbonate percentage chemistry calculation. need help urgently?

  • HI, could someone please help solve and explain this chemistry calculation to me. Thanks! The problem is: 1.00gram of a sample of limestone was dissolved in dilute hydrochloric acid and then ammonium oxalate and ammonium hydroxide added. The precipitate of calcium oxalate was filtered and dried and found to weigh 0.448grams. Calculate the percentage calcium carbonate in the limestone. Note: the answer is 35% but I really need someone to SHOW how to get this because I keep getting it wrong. Thanks again!

  • Answer:

    The reactions are CaCO3 + 2 HCl = CaCl2 + CO2 + H2O CaCl2 + (NH4)2C2O4 = CaC2O4 + 2 NH4Cl moles CaC2O4 = 0.448 g/ 128.096 g/mol=0.00350 the ratio between CaC2O4 and CaCO3 is 1 : 1 moles CaCO3 = 0.00350 mass CaCO3 = 0.00350 mol x 100 g/mol=0.350 g % CaCO3 = 0.350 x x 100 / 1.00 g= 35.0

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The reactions are CaCO3 + 2 HCl = CaCl2 + CO2 + H2O CaCl2 + (NH4)2C2O4 = CaC2O4 + 2 NH4Cl moles CaC2O4 = 0.448 g/ 128.096 g/mol=0.00350 the ratio between CaC2O4 and CaCO3 is 1 : 1 moles CaCO3 = 0.00350 mass CaCO3 = 0.00350 mol x 100 g/mol=0.350 g % CaCO3 = 0.350 x x 100 / 1.00 g= 35.0

Dr.A

with all due respect, :P this isn't such a difficult question.CaCO3 + 2HCl gives CaCl2 + water plus CO2... then the calcium chloride obtained is reacted with ammonium oxalate and ammonium hydroxide...so we get 1 mol of calcium oxalate from 1 mol of calcium chloride..... therefore, (i dnt kno which law this is...lol) using mass conservation, 111g (1 mol of CaCl2) give 128 g of calcium oxalate.. thus....0.448 g of calcium oxalate is given by 111*0.448/128=0.3885g therefore the calcium chloride reqd. is 0.3885 g.... but 100 g of calcium carbonate gives 111 g of CaCl2 thus x g of CaCO3 gives 0.3885 g of CaCl2 here x = 0.3885*100/111=0.35 g therefore the given 1 g of sample contains 0.35 g of CaCO3... thus percentage of calcium carbonate is 0.35*100/1=35 % hope this helps you :)

adidela

You are producing calcium oxalate from calcium carbonate. Via a roundabout route but that is of no consequence. Write a balanced equation CaCO3 + (COOH)2 → (COO)2Ca + H2O +CO2 1mol CaCO3 reacts with 1mol (COOH)2 to produce 1mol (COO)2Ca Molar mass (COO)2Ca = 128.0977 g/mol 0.448g = 0.448/128.0977 = 0.003497mol This must come from 0.003497 mol CaCO3 Molar mass CaCO3 = 100.0875 g/mol 0.003497 mol = 100.0875*0.003497 = 0.350g %CaCO3 in sample = 0.35/1*100 = 35%

Trevor H

You are producing calcium oxalate from calcium carbonate. Via a roundabout route but that is of no consequence. Write a balanced equation CaCO3 + (COOH)2 → (COO)2Ca + H2O +CO2 1mol CaCO3 reacts with 1mol (COOH)2 to produce 1mol (COO)2Ca Molar mass (COO)2Ca = 128.0977 g/mol 0.448g = 0.448/128.0977 = 0.003497mol This must come from 0.003497 mol CaCO3 Molar mass CaCO3 = 100.0875 g/mol 0.003497 mol = 100.0875*0.003497 = 0.350g %CaCO3 in sample = 0.35/1*100 = 35%

Trevor H

with all due respect, :P this isn't such a difficult question.CaCO3 + 2HCl gives CaCl2 + water plus CO2... then the calcium chloride obtained is reacted with ammonium oxalate and ammonium hydroxide...so we get 1 mol of calcium oxalate from 1 mol of calcium chloride..... therefore, (i dnt kno which law this is...lol) using mass conservation, 111g (1 mol of CaCl2) give 128 g of calcium oxalate.. thus....0.448 g of calcium oxalate is given by 111*0.448/128=0.3885g therefore the calcium chloride reqd. is 0.3885 g.... but 100 g of calcium carbonate gives 111 g of CaCl2 thus x g of CaCO3 gives 0.3885 g of CaCl2 here x = 0.3885*100/111=0.35 g therefore the given 1 g of sample contains 0.35 g of CaCO3... thus percentage of calcium carbonate is 0.35*100/1=35 % hope this helps you :)

adidela

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