Question

According to the $CaC{{O}_{3}}+HCl\to CaC{{l}_{2}}+C{{O}_{2}}+{{H}_{2}}O$. What mass of $CaC{{O}_{3}}$ is required to react completely with 25 ml. of 0.75M HCl?

Answer 1

Hint: As we know that Molarity is defined as the number of moles of solute per litre of solution. The unit of molarity is M. The formula of molarity is given as:
$Molarity=\dfrac{number\text{ }of\text{ }moles\text{ }of\text{ }solute}{volume\text{ }of\text{ }solution\text{ }in\text{ }litres}$

Complete Step by step solution:
- As we know that from the definition of molarity, the number of moles are calculated for 1 litre or 1000 ml of the solution.
- So, 0.75 M of HCl = 0.75 mol of HCl $\times$ the molecular weight of HCl that is dissolved in 1000 ml of water.
- Or we can say that HCl is present in that of 1 L of water.
= 27.375 g of HCl is present in that of 1 L of water.
- Therefore, we can say that 1000 ml of solution contains 27.375 g of HCl. Or 1 ml of solution has$\dfrac{27.375}{1000}\times 1$ HCl.
- According to the given chemical equation, $CaC{{O}_{3}}+HCl\to CaC{{l}_{2}}+C{{O}_{2}}+{{H}_{2}}O$
-2 mol of (2$\times$36.5=71 g) HCl will react with 1 mol of (100 g) of $CaC{{O}_{3}}$.
- Therefore, we can say that the amount of $CaC{{O}_{3}}$that will react with 0.6844 g =
 $\begin{align}
 &\implies \dfrac{100}{71}\times 0.6844\text{ }g \\
& \implies0.9639\text{ }g \\
\end{align}$

- Hence, we can conclude that 0.9639 g mass of$CaC{{O}_{3}}$ is required to react completely with 25 ml. of 0.75M HCl.

Note: - We should not get confused in terms of normality and molarity. Normality is the number equivalent of solute dissolved per litre of solution. And, Molarity is defined as the number of moles of solute per litre of solution. The unit of normality is N. The unit of molarity is M.