Question 6b9a2

May 25, 2017

$\text{a.}$ $0.02$ $\text{mol}$

$\text{b.}$ $0.89$ $\text{g}$

Explanation:

In order to answer these questions we first need to know the equation representing this chemical reaction.

In this case, it is a reaction between $\text{HCl}$ and ${\text{Ca(OH)}}_{2} :$

$\text{Ca(OH)}$2 $+$ $\text{HCl}$ $\rightarrow$ $\text{CaCl}$2 $+$$\text{H}$2$\text{O}$

Let's balance the equation:

$\text{Ca(OH)}$2 $+$ 2$\text{HCl}$ $\rightarrow$ $\text{CaCl}$2 $+$ 2$\text{H}$2$\text{O}$

$\text{a.}$ Using the formula $n = \frac{m}{M} :$

$R i g h t a r r o w n \left({\text{Ca(OH)") = frac(0.75 " g")(74.10 " g mol}}^{- 1}\right)$

$R i g h t a r r o w n \left(\text{Ca(OH)}\right) = 0.01$ $\text{mol}$

In the balanced chemical equation above, the stoichiometric ratio between $\text{HCl}$ and $\text{Ca(OH)}$ is $2 : 1$.

$\therefore n \left(\text{HCl}\right) = 2 \times 0.01$ $\text{mol}$

$R i g h t a r r o w n \left(\text{HCl}\right) = 0.02$ $\text{mol}$

$\text{b.}$ Let's set up a ratio using the molecular masses of $\text{Ca(OH)}$ and $\text{CaCl}$:

Rightarrow frac("MM of Ca(OH)")("MM of CaCl") = frac("m of Ca(OH)")("m of CaCl")#

$R i g h t a r r o w \frac{74.10 \text{ g mol"^(- 1))(110.98 "g mol"^(- 1)) = frac(0.75 " g}}{x}$

We need to find $x$, which is the mass of $\text{CaCl}$ formed in the reaction:

$R i g h t a r r o w 0.6676878717 = \frac{0.75 \text{ g}}{x}$

$\therefore x = 0.89$ $\text{g}$

Therefore, the number of moles of $\text{HCl}$ needed to completely react with ${\text{Ca(OH)" }}_{2}$ is $0.02$ $\text{mol}$, while the mass of $\text{CaCl}$ formed in the reaction is $0.89$ $\text{g}$.