Answer to Question #75957 in General Chemistry for Mary

Question

A 1.433 gram sample o a calcium carbonate mixture is treated with HCl giving off carbon dioxide gas. After the reaction the mixture has a mass of 1.194 grams.
A) How many moles of CO2 are evolved? (SHOW ALL WORK)
B) How many grams of CaCO3 are in the mixture? (SHOW ALL WORK)
C) What is the % CaCO3 in the mixture? (SHOW ALL WORK)

Answer 1

Answer:
CaCO3 + 2HCl = CaCl2 + CO2 + H2O
1.433 - 1.194 = 0.239 g of CO2 evolved
0.239 / 44 g/mol = 0.00543 moles of CO2 evolved = moles CaCO3
Molar mass = 100 g/mol
100 x 0.00543 = 0.543 g of CaCO3
% = 0.543 x 100 / 1.433 = 37.9