Calcium carbonate reacts with aqueous HCl to give CaCl2 and CO2 according to the reaction given below

CaCO3(s)+2HCl(aq)⟶CaCl2(aq)+CO2(g)+H2O(l)

what mass of CaCO3 is required to react completely with 25 mL of 0.75 M HCl?

1) 0.6844

2) 0.9375

3) 0.4265

4) 0.2785

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Solution

The given equation is
CaCO3+2HCl --> CaCl2 + H2O + CO2
So the mole ratio between calcium carbonate and HCl is 1:2
Number of moles of HCl = molarity x volume
= 0.75x0.025 Litres = 0.01875 moles
So the number of moles of Calcium carbonate should be half of number of moles of HCl
Number of moles of CaCO3 = 1/20.01875=0.009375
Molar mass of CaCO3= 100u
Mass of calcium carbonate = moles molar mass = 0.009375*100= 0.9375g
So option (2) is correct

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Calcium carbonate reacts with aqueous HCl to give CaCl2 and CO2 according to the reaction given below CaCO3(s)+2HCl(aq)⟶CaCl2(aq)+CO2(g)+H2O(l) what mass of CaCO3 is required to react completely with 25 mL of 0.75 M HCl? 1) 0.6844 2) 0.9375 3) 0.4265 4) 0.2785

Calcium carbonate reacts with aqueous HCl to give CaCl2 and CO2 according to the reaction given below

CaCO3(s)+2HCl(aq)⟶CaCl2(aq)+CO2(g)+H2O(l)

what mass of CaCO3 is required to react completely with 25 mL of 0.75 M HCl?

1) 0.6844

2) 0.9375

3) 0.4265

4) 0.2785