# Question #c882e

Oct 3, 2016

$\textsf{\left(a\right)}$

$\textsf{\left[H C l\right] = 0.5056 \textcolor{w h i t e}{x} \text{mol/l}}$

$\textsf{\left(b\right)}$

$\textsf{\left[R b O H\right] = 0.9118 \textcolor{w h i t e}{x} \text{mol/l}}$

#### Explanation:

$\textsf{\left(a\right)}$

$\textsf{B a {\left(O H\right)}_{2} + 2 H C l \rightarrow B a C {l}_{2} + 2 {H}_{2} O}$

$\textsf{c = \frac{n}{v}}$

$\therefore$$\textsf{n = c \times v}$

$\therefore$$\textsf{{n}_{B a {\left(O H\right)}_{2}} = 0.1529 \times 43.09 = 6.5884 \textcolor{w h i t e}{x} \text{mmol}}$

From the equation we can see that the no. moles of $\textsf{H C l}$ must be twice this.

$\therefore$$\textsf{{n}_{H C l} = 6.5884 \times 2 = 13.177 \textcolor{w h i t e}{x} \text{mmol}}$

$\therefore$$\textsf{\left[H C l\right] = \frac{c}{v} = \frac{13.177 \times \cancel{{10}^{- 3}}}{\frac{26.06}{\cancel{1000}}} = 0.5056 \textcolor{w h i t e}{x} \text{mol/l}}$

$\textsf{\left(b\right)}$

$\textsf{H C l + R b O H \rightarrow R b C l + {H}_{2} O}$

$\textsf{{n}_{H C l} = c \times v = 0.5056 \times 27.05 = 13.676 \textcolor{w h i t e}{x} \text{mmol}}$

From the equation we can see that the no. moles of $\textsf{R b O H}$ must be the same .

$\therefore$$\textsf{{n}_{R b O H} = 13.676 \textcolor{w h i t e}{x} \text{mmol}}$

$\therefore$$\textsf{\left[R b O H\right] = \frac{c}{v} = \frac{13.676 \times \cancel{{10}^{- 3}}}{\frac{15.00}{\cancel{1000}}} = 0.9118 \textcolor{w h i t e}{x} \text{mol/l}}$