For the rxn CaCo3(s)+2Hcl(aq)_CaCl(aq)+Ca2(g)+H 2O(l)68.1 g solid CaCo3 is mixed with 51.6Hcl what number of grams of Co2 will be produced

29.9 g of CO₂ will be produced by the reaction.

Explanation:

This is the reaction:

CaCO₃(s)  + 2HCl (aq)  →  CaCl₂(aq)  + CO₂(g) +  H₂O(l)

First of all, we state the moles of each reactant:

68.1 g . 1mol/ 100.08g = 0.680 mol of carbonate

51.6 g . 1 mol/36.45g = 1.46 mol of acid.

The solid salt is the limiting reactant. Ratio is 2:1

2 moles of acid can react to 1 mol of salt

1.46 mol of acid may react with (1.46 . 1)/2 = 0.727 moles

As we only have 0.680 moles of salt, we do not have enough.

Let's work at the product side. Ratio is 1:1

1 mo of salt can produce 1 mol of gas

0.680 moles will produce 0.680 moles of gas

We convert the moles to mass → 0.680 mol . 44g / 1mol = 29.9 g of CO₂

i think it is true since you can test the answer and can use observations to complete it.