Calcium carbonate reacts with aqueous HCl to give CaCl2 and CO2 according to the reaction, CaCO3(s) + 2HCl (aq) -- CaCl2(aq) + CO2(g) +H2O(l)

  • What mass of CaC03 is required to react completely with 25 mL of 0.75 M HCl ?

From the reaction it is clear that 1 mol of CaCO3 reacts with 2 moles of HCl.

Amount of HCl in 25ml 0.75M can be calculated as
1 mol of HCl = 36.5 g in 1000 ml
Therefore 0.75 M = 27.38 g in 1000 ml

Therefore in 25 ml it will be 27.38/40 =0.6845 g = 0.01875 mol
The ratio of reactants = 1:2
Hence Moles of Ca-carbonate needed for 0.01875 moles = 0.009375 mol

Mass = 0.009375 X 100 = 0.9375 gms
 

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