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Answer 1

Answer: 0.9375 g

Explanation:

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

$\text{Moles of solute}={\text{Molarity of the solution}}\times{\text{Volume of solution (in L)}}$     .....(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:

$\text{Moles of} HCl={0.75}\times{0.025}=0.01875moles$  

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

According to stoichiometry :

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

Thus 0.01875 moles of $HCl$ will require=$\frac{1}{2}\times 0.01875=0.009375moles$  of $CaCO_3$

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

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