If 3.00 g of limestone reacted, what mass of calcium chloride would be produced?

a) calcium chloride
b) carbon dioxide
c) water

A) CaCO₃ + 2 HCl = CaCl₂ + H₂O + CO₂
  ↓                            ↓
1 mole                 1 mole

molar mass :

CaCO₃ = 100.0 g/mol     CaCl₂ = 111 g/mol

 100 g CaCO₃ 111 g caCl₂
  3.00 g CaCO₃ ?

3.00 * 111 / 100 = 

333 / 100 = 3.33 g of CaCl₂


b) molar mass CO₂ = 44.0 g/mol

100 g CaCO₃ 44.0 g CO₂
3.00 g CaCO3 ?

3.00 * 44 / 100 = 

132 / 100 = 1.32 g of CO2


c) molar mass H₂O = 18.0 /mol

100 g CaCO3 18.0 g 
3.00 g CaCO3 ?

3.00 * 18.0 / 100 =

54 / 100 = 0.54 g of H₂O


hope this helps!

Hey how did you manage to type such a big lab report