Question

Calculate the volume (in mL) of 0.620 M HNO3 needed to react completely with 5.31 g of CaCO3 in a gas-forming reaction?

Answer 1

CaCO3 + 2 HNO3 = Ca(NO3)2 + H2O + CO2

Number of moles of CaCO3 = amount of CaCO3 (g)/ mw of CaCO3 (g/mol)

                                                  = 5.31 g/ 100.09 g/mol= 0.0531 moles

According to the above balanced equation

2 moles of HNO3 reacts with 1 mole of CaCO3

Therefore 0.0531 moles of CaCO3 will require = 2 X .0531 moles of HNO3= 0.1062 moles of HNO3

Molarity of HNO3 X Volume of HNO3 (L) = number of moles of HNO3 required

0.62 M X Volume of HNO3 (L) = 0.1062 mole

Volume of HNO3 (L)= 0.1062/0.62= 0.17114 L

Volume of HNO3= 171.14 mL