Answer:
22.7 g of CaCl₂ are produced in the reaction
Explanation:
This is the reaction:
CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O
Now, let's determine the limiting reactant.
Let's divide the mass between the molar mass, to find out moles of each reactant.
29 g / 100.08 g/m = 0.289 of carbonate
15 g / 36.45 g/m = 0.411 of acid
1 mol of carbonate must react with 2 moles of acid
0.289 moles of carbonate will react with the double of moles (0.578)
I only have 0.411 of HCl, so the acid is the limiting reactant.
Ratio is 2:1, so I will produce the half of moles, of salt.
0.411 / 2 = 0.205 moles of CaCl₂
Mol . molar mass = mass → 0.205 m . 110.98 g/m = 22.7 g
