Question
Calcium carbonate reacts with aqueous HCl to give CaCl2 and CO2 according to the reaction,
What mass of CaCO3 is required to react completely with 25 mL of 0·75M HCl?
Solution
1000 mL of HCl contains 0·75 mol
= 0.75 x 36.5
= 27.375 g
25 mL of HCl contains 
= 0.684 g
The reaction is
CaCO3(s) + 2HCl(aq)
CaCl2(aq) + CO2(g) + H2O(l)
1 mol 2 mol
40+12+48 2 x 36.5
= 100 g = 73 g
? 0.684 g
Now 73 g of HCl are required to react with 100g of CaCO3
0.684 g of HCl are required to react with
= 0.75 x 36.5
= 27.375 g
25 mL of HCl contains = 0.684 g
The reaction is
CaCO3(s) + 2HCl(aq)
CaCl2(aq) + CO2(g) + H2O(l)1 mol 2 mol
40+12+48 2 x 36.5
= 100 g = 73 g
? 0.684 g
Now 73 g of HCl are required to react with 100g of CaCO3
0.684 g of HCl are required to react with
