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?
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 0.684g of HCl are required to react with
52 u He 
25 mL of HCl contains 
CaCl2(aq) + CO2(g) + H2O(l)
0.684 g of HCl are required to react with
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