How to find out the resulting weight of an acid and metal carbonate reaction?

Question

If I reacted 66ml of 0.18M HCl with 1g of baking soda, the stoichiometry ratio should be around 1:1.

I want to know how much CO2 is produced in the reaction (to find out the remaining weight of whats left).

Also in addition to that, it'd be helpful to know what would happen if there was 2g of baking soda, making the ratio to 1:2 (HCl:NaHCO3)

Answer 1

Let's look at the reaction:

HCl + NaHCO₃ ==> NaCl + H₂O + CO₂

To find mass of CO₂ formed, we will first convert to moles, then back to grams:

moles HCl used = 66 ml x 1 L / 1000 ml x 0.18 mol / L = 0.012 mols HCl

mols NaHCO₃ used = 1 g x 1 mol / 84.0 g = 0.012 mols NaHCO₃

Since they are both present in the same molar amounts, and since they react in a 1:1 mol ratio, neither will be limiting and we can now determine moles and grams of CO₂ produced:

mols CO₂ produced = 0.012 mols reactant x 1 mol CO₂ / mol reactant = 0.012 mols CO₂ produced

grams CO₂ = 0.012 mols CO₂ x 44 g CO₂ / mol = 0.53 g CO₂ produced

If there were 2 g of NaHCO₃, the limiting reactant would then be HCl, and the mols of HCl would thus determine mols and grams of CO₂ produced. So, the answer would be the same, i.e. 0.53 g CO₂ produced.

Please note that the stoichiometric ratio of HCl : NaHCO₃ is always 1:1 and not 1:2. The 1:2 ratio is only because you added twice as much NaHCO₃, but that's not the ratio in the BALANCED EQUATION, which is what's important.