Question

Calcium carbonate reacts with aqueous HCl to give CaCl₂ and CO₂ according to the reaction, CaCO₃ (s) + 2 HCl (aq) → CaCl₂ (aq) + CO₂ (g) + H₂O(l) What mass of CaCO₃ is required to react completely with 25 mL of 0.75 M HCl?​

Answer 1

Answer:

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Answer 2

Answer:

Let us first calculate the mass of HCl that is present in 25 mL of 0.75 M HCl solution. Molarity of a solution is given by the following relation

Molarity = number of moles of solute / Volume of solution (in litres)

0.75 = n / 0.025

This gives n = 0.01875

Now, number of moles is given by the following realtion

number of moles = mass of substance / molar mass of substance

So, the mass of HCl will be

0.01875 = mass of HCl / 36.5

mass of HCl = 0.684g

Now, a balanced chemical equation showing the reaction of HCl and CaCO3 is as follows

CaCO3 + 2HCl → CaCl2 + H2O + CO2

According to the above equation, 100g of CaCO3 is required to react completely with 71g of HCl. That is

73g of HCl → 100g of CaCO3

1g of HCl → (100 / 73) g of CaCO3

So, 0.684g would require = (100 / 73) X 0.684 g of CaCO3

= 0.937g of CaCO3

Thus 0.937g of CaCO3 would be required to completely react with 25ml of 0.75M HCl.