Question

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 ?

Answer 1

0.75 M solution of HCl = 0.75 moles of HCl in 1 L water
Number of moles of HCl in 25mL of 0.75 M solution = (0.75/1000)25 =0.019 moles
From equation it is clear that one mole CaCO3 reacts with two moles of HCl
So,
Number of moles of CaCO3 required to react with 0.019 moles of HCl = (1/2)0.019 = 0.001
Mass of 0.001 mole CaCO3 = (0.001 )(Molar mass of CaCO3) =(0.001 )( 100) = 0.1g




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Answer 2

Answer: :/

Explanation:

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