Question

Consider the following reaction:

2HCl+CaCO3→CaCl2+H2O+CO2

How many mols of calcium chloride can be produced if you begin with 13.15 mL of 0.62 M HCl and 10.32 grams of calcium carbonate? Record your answer in scientific notation, using 3 significant figures.

Answer 1

Number of moles of HCl = molarity * volume of solution in L

Number of moles of HCl = 0.62 * 0.01315 = 0.00815 mole

Number of moles of CaCO3 = 10.32 g / 100.0869 g/mol = 0.103 mol

From the balanced equation we can say that

2 mole of HCl requires 1 mole of CaCO3 so

0.00815 mole of HCl will require

= 0.00815 mole of HCl *(1 mole of CaCO3 / 2 mole of HCl)

= 0.00408 mole

But we have 0.103 mole of CaCO3 which is in excess so CaCO3 is an excess reactant and HCl is limiting reactant

From the balanced equation we can say that

2 mole of HCl produces 1 mole of CaCl2 so

0.00815 mole of HCl will produce

= 0.00815 mole of HCl *(1 mole of CaCl2 / 2 mole of HCl)

= 0.00408 mole of CaCl2

mass of 1 mole of CaCl2 = 110.98 g so

the mass of 0.00408 mole of CaCl2 = 0.453 g

Therefore, the mass of CaCl2 produced would be 0.453 g