Answer :
Answer: 0.9375 g
Explanation:
To calculate the number of moles for given molarity, we use the equation:
[tex]\text{Moles of solute}={\text{Molarity of the solution}}\times{\text{Volume of solution (in L)}}[/tex] .....(1)
Molarity of [tex]HCl[/tex] solution = 0.75 M
Volume of [tex]HCl[/tex] solution = 25.0 mL = 0.025 L
Putting values in equation 1, we get:
[tex]\text{Moles of} HCl={0.75}\times{0.025}=0.01875moles[/tex]
[tex]CaCO_3(s)+2HCl(aq)\rightarrow CaCl_2(s)+CO_2(g)+H_2O(l)[/tex]
According to stoichiometry :
2 moles of [tex]HCl[/tex] require = 1 mole of [tex]CaCO_3[/tex]
Thus 0.01875 moles of [tex]HCl[/tex] will require=[tex]\frac{1}{2}\times 0.01875=0.009375moles[/tex] of [tex]CaCO_3[/tex]
Mass of [tex]CaCO_3=moles\times {\text {Molar mass}}=0.009375moles\times 100g/mol=0.9375g[/tex]
Thus 0.9375 g of [tex]CaCO_3[/tex] is required to react with 25.0 ml of 0.75 M HCl