Answer:
19.7 g of CaCl₂ are produced
Explanation:
This is the reaction:
CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O
As we have the mass of reactants, we should determine the moles we used, of each
29g . 1mol / 10.08 g = 0.289 moles
13 g . 1mol / 36.45 g = 0.356 moles
Ratio is 1:2, so If I have 1 mol of carbonate I need the doulbe of hydrochloric. Certainly the limiting reactant is the HCl.
I have 0.289 moles of salt and I need 0.578 (x2) moles of acid. - I only have 0.356 moles.
As we found out the limiting reactant we can work with the reaction. Ratio with CaCl₂ is 2:1 so,
2 moles of HCl produce 1 mol of CaCl₂
Then, 0.356 moles of HCl would produce (0.356 . 1) / 2 = 0.178 moles.
To convert the moles to mass we do: 0.178 mol . 110.98 g / 1mol = 19.7 g
