CaCO3 (s) + 2HCl (aq) → CaCl2(aq) + CO2(g) + H2O(l)
What mass of CaCl2 will be formed when 250 mL of 0.76 M HCl reacts with 1000 g of CaCO3? Name the limiting reagent. Calculate the number of moles of CaCl2 formed in the reaction.
Calculation:
The number of moles of HCl taken = MV/1000
The number of moles of HCl taken = 0.76*250/1000
The number of moles of HCl taken = 0.19
Number of moles of CaCO3 = Mass/Molar mass
Number of moles of CaCO3 = 1000/100
Number of moles of CaCO3 = 10
- If CaCO3 is completely consumed,
1 mol of CaCO3 = 1 mol CaCl2
10 mol CaCO3 = 10mol CaCl2
- If HCl is completely consumed,
2 mol HCl = 1 mol CaCl2
0.19mol HCl = ½ × 0.19 mol CaCl2
0.19mol HCl = 0.095 mol CaCl2
HCl will be the limiting reagent
Hence, the number of moles of CaCl2 formed will be 0.095mol