Answer:
19.7 g of CaCl₂ are produced in the reaction
Explanation:
In this excersise we need to know the limiting reactant in order to determine the mass produced of a one of the products
The reaction is: CaCO₃ (s) + 2HCl (aq) → CaCl₂(aq) + H₂O (l) + CO₂(g)
First of all we must find the limiting reactant. For that purpose, we convert the mass of reactants to moles
25 g / 100.08 g/mol = 0.249 moles of carbonate
13 g / 36.45 g/mol = 0.357 moles of HCl
We work with the stoichiometry of the reaction:
1 mol of carbonate reacts with 2 moles of hydrochloric
Then, 0.249 moles of carbonate must react with (0.249 . 2) /1 = 0.498 moles of HCl (We do not have enough HCl, so this is the limtiing reactant)
We work with the stoichiometry reactant / product
2 moles of HCl can produce 1 mol of CaCl₂
Therefore 0.357 moles of HCl must produce (0.357 .1) / 2 = 0.178 moles of chloride.
We convert the moles to mass → 0.178 mol . 110.98 g /1mol = 19.7g
