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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)

0.9375 g

Explanation:

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

    .....(1)

Molarity of solution = 0.75 M

Volume of solution = 25.0 mL = 0.025 L

Putting values in equation 1, we get:

According to stoichiometry :

2 moles of require = 1 mole of

Thus 0.01875 moles of will require=  of

Mass of

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

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