15.2 g
Begin by calculating the molar mass of the reactants and the desired product. First, lookup the atomic weights of each element involved.
Calcium = 40.078
Carbon = 12.0107
Oxygen = 15.999
Hydrogen = 1.00794
Chlorine = 35.453
Now sum up the products of the atomic weights of each element times the number of times each element is used in each compound
Calcium Chloride (CaCl2)
40.078 + 2 * 35.453 = 110.984
Calcium Carbonate (CaCO3)
40.078 + 12.0107 + 3 * 15.999 = 100.0857
Hydrochloric acid (HCl)
1.00794 + 35.453 = 36.46094
Figure out how many moles of each reactant is available by dividing mass by molar mass
Calcium Carbonate (CaCO3)
29.0 g / 100.0857 g/mol = 0.289752 mol
Hydrochloric acid (HCl)
10.0 g / 36.46094 g/mol = 0.274266 mol
Create a balanced equation for the reaction
CaCO3 + 2 HCl => CaCl2 + CO2 + H2O
Looking at the balanced equation, it takes 2 moles of HCl for each mole of CaCO3, since the number of available moles for each reactant is about equal, the limiting reactant will be HCl. So for each mole of HCl, 0.5 moles of CaCO3 will be needed.
0.5 * 0.274266 = 0.137133
So we'll be producing 0.137133 moles of CaCl2. Just multiply by the previously calculated molar mass
0.137133 mol * 110.984 g/mol = 15.21956887 g
Since we only have 3 significant digits available in our measurements, round the result to 3 significant digits.
15.21956887 g = 15.2 g
