CaCO₃( s)+2 HCl(aq) → CaCl₂(aq)+CO₂( g)+H₂ O(l)
Consider the above reaction, what mass of CaCl₂ will be formed if 250 mL of 0.76 M HCl reacts with 1000 g of CaCO₃ ?
(Given : Molar mass of Ca, C, O, H and Cl are 40,12,16,1 and 35.5 g mol⁻¹, respectively)

Solution

Chemistry ma’am · 4 steps

Turn both reactants into moles first; the tiny amount of HCl runs out long before 1000 g of chalk does.

1. Moles of each. HCl: 0.250 L × 0.76 mol L⁻¹ = 0.19 mol. CaCO₃ (molar mass 40 + 12 + 48 = 100 g mol⁻¹): 1000/100 = 10 mol.
2. Find the limiting reagent. The equation needs 2 mol HCl for every 1 mol CaCO₃. To use up 10 mol CaCO₃ you would need 20 mol HCl, but only 0.19 mol is there. So HCl is the limiting reagent, and CaCO₃ is in large excess.

   n_(CaCl₂) = n_(HCl)/2 = 0.19/2 = 0.095 mol

3. Convert to mass. Molar mass of CaCl₂ is 40 + 2 × 35.5 = 111 g mol⁻¹. Mass = 0.095 × 111 = 10.545 g.
4. So. 10.545 g of CaCl₂ forms, which is option (c).

The idea to keep. Never trust the bigger mass to tell you the limiting reagent. Convert everything to moles, divide by the coefficients, and let the smaller one decide the product.

Answer from the NTA final answer key. Solution by PrepWiser.