Question

20g of 40% pure CaCO3 reacts with 5g of HCL to produce CaCl2, H2O and CO2.

Find which one is limiting reagent and why?

Answer 1

To determine the limiting reagent, compare the moles of each reactant to the stoichiometric ratio in the balanced equation. In this case, the HCl is the limiting reagent because its mole ratio is greater than 1. CaCO3 is in excess and there will be some leftover after the reaction.

Step-by-step explanation:

To determine the limiting reagent, we need to compare the moles of each reactant to the stoichiometric ratio given in the balanced equation. The balanced equation is:

CaCO3 + 2HCl -> CaCl2 + H2O + CO2

Let's calculate the number of moles of each reactant:

20g CaCO3 x (1mol CaCO3/100g CaCO3) = 0.2 mol CaCO3

5g HCl x (1mol HCl/36.5g HCl) = 0.14 mol HCl

According to the balanced equation, 1 mol of CaCO3 reacts with 2 mol of HCl. Therefore, the mole ratio is 0.2 mol CaCO3 : 0.4 mol HCl.

Since the mole ratio is greater than 1, HCl is the limiting reagent. This means that CaCO3 is in excess and there will be some leftover after the reaction is complete.