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1. Calculate the mass of CaCO3(s) required to react with 25.0 mL of 0.75M HCl(aq).
CaCO3(s) + 2 HCl(aq) → CaCl2(aq) + CO2(g) + H2O(1)

Answer 1

Answer: 0.9375 g

Step-by-step explanation:

To calculate the number of moles for given molarity, we use the equation:

`\text{Moles of solute}={\text{Molarity of the solution}}*{\text{Volume of solution (in L)}}` .....(1)

Molarity of
`HCl` solution = 0.75 M

Volume of
`HCl` solution = 25.0 mL = 0.025 L

Putting values in equation 1, we get:

`\text{Moles of} HCl={0.75}*{0.025}=0.01875moles`

`CaCO_3(s)+2HCl(aq)\rightarrow CaCl_2(s)+CO_2(g)+H_2O(l)`

According to stoichiometry :

2 moles of
`HCl` require = 1 mole of
`CaCO_3`

Thus 0.01875 moles of
`HCl` will require=
`(1)/(2)* 0.01875=0.009375moles` of
`CaCO_3`

Mass of
`CaCO_3=moles* {\text {Molar mass}}=0.009375moles* 100g/mol=0.9375g`

Thus 0.9375 g of
`CaCO_3` is required to react with 25.0 ml of 0.75 M HCl