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

The correct option is 0.9639 g.

To find the mass of calcium carbonate (${\mathrm{CaCO}}_3$) required to react completely with the given volume and concentration of hydrochloric acid ($\mathrm{HCl}$), we can follow these steps:

Step 1: Understand the balanced chemical equation
The balanced chemical equation for the reaction is:
${\mathrm{CaCO}}_3\left(s\right)+2\mathrm{HCl}\left(\mathrm{aq}\right)\to {\mathrm{CaCl}}_2\left(\mathrm{aq}\right)+{\mathrm{CO}}_2\left(g\right)+H_{2}O\left(l\right)$
From the stoichiometry of the reaction, we can see that 1 mole of ${\mathrm{CaCO}}_3$ reacts completely with 2 moles of $\mathrm{HCl}$.

Step 2: Calculate the number of moles of HCl
We are given:
Volume of $\mathrm{HCl}$ solution = 25 mL = $0.025\text{L}$
Molarity (M) of $\mathrm{HCl}$ solution = 0.75 M = $0.75\text{mol/L}$
Using the formula for molarity:
$\text{Moles of HCl}=\text{Molarity}\times \text{Volume (in L)}$
$\text{Moles of HCl}=0.75\times 0.025=0.01875\text{mol}$

Step 3: Calculate the moles of CaCO₃ required
According to the balanced equation, the mole ratio of ${\mathrm{CaCO}}_3$ to $\mathrm{HCl}$ is 1:2. Therefore:
$\text{Moles of}{\mathrm{CaCO}}_3=\frac{\text{Moles of HCl}}{2}$
$\text{Moles of}{\mathrm{CaCO}}_3=\frac{0.01875}{2}=0.009375\text{mol}$

Step 4: Calculate the molar mass of CaCO₃
Using the standard atomic masses (Ca = 40.08 g/mol, C = 12.01 g/mol, O = 16.00 g/mol):
$\text{Molar mass of}{\mathrm{CaCO}}_3=40.08+12.01+(3\times 16.00)=100.09\text{g/mol}$

Step 5: Convert moles of CaCO₃ to mass
Using the formula:
$\text{Mass}=\text{Moles}\times \text{Molar mass}$
$\text{Mass of}{\mathrm{CaCO}}_3=0.009375\text{mol}\times 100.09\text{g/mol}\approx 0.9639\text{g}$
Therefore, the mass of calcium carbonate required to react completely with the given hydrochloric acid solution is 0.9639 g.