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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)

User BigBalli
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1 Answer

6 votes

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

User Mahbub Tito
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