Answer to Question #59869 in General Chemistry for Tim

Question #59869

Could you confirm the following is correct?
How much gypsum is needed to neutralize 8.3 grams of urea?
First need to calculate how much ammonium carbonate is created from 8.3 grams of Urea.
NH2CONH2 + H2O  (NH4)2CO3
Urea Water Ammonium Carbonate

8.3 g of urea / 60.06 g per mol = 0.1382 mol of urea

0.1382 mol of (NH4)2CO3 x 96.1 g (NH4)2CO3 / 1 mol (NH4)2CO3 = 13.3 g (NH4)2CO3

Next, need to calculate the amount of gypsum needed to neutralize 13.3 g of (NH4)2CO3

(NH4)2CO3 + CaSO4 2H2O  CaCO3 + (NH4)2SO4
Ammonium Carbonate gypsum calcium carbonate ammonium sulphate

13.3 g of (NH4)2CO3 / 96.1 g per mol = 0.1384 mol of (NH4)2CO3

0.1384 mol of CaSO4 2H2O x 172.17 g CaSO4 2H2O / 1 mol CaSO4 2H2O = 23.8 g CaSO4 2H2O

Expert's answer

The equation must be:
NH2CONH2 + 2H2O=(NH4)2CO3
But this additional coefficient does not influence further calculations.
All the rest calculations are right.