# Question add6d

May 7, 2017

The mass of $\text{NaBr}$ will be 23.4 g.

#### Explanation:

If there is such a process, the balanced equation will probably be

${M}_{\textrm{r}} : \textcolor{w h i t e}{m} 187.77 \textcolor{w h i t e}{m m m m l} 102.89$
color(white)(mmm)n"AgBr" + ? → n"NaBr" + ?#

where $n$ is the same number on each side of the equation.

Step 2. Calculate the moles of $\text{AgBr}$.

$\text{Moles of AgBri" = 42.7 color(red)(cancel(color(black)("g AgBr"))) × ("1 mol AgBr")/(187.7 color(red)(cancel(color(black)("g AgBr")))) = "0.2275 mol AgBr}$

Step 3. Calculate the moles of $\text{NaBr}$

$\text{Moles of NaBr" = 0.2275 color(red)(cancel(color(black)("mol AgBr"))) × (stackrelcolor(blue)(1)(color(red)(cancel(color(black)(n)))) "mol NaBr")/(color(red)(cancel(color(black)(n))) color(red)(cancel(color(black)("mol AgBr")))) = "0.2275 mol NaBr}$

4. Calculate the mass of $\text{NaBr}$.

$\text{Mass of H"_2"O" = 0.2275 color(red)(cancel(color(black)("mol NaBr"))) × "102.89 g NaBr"/(1 color(red)(cancel(color(black)("mol NaBr")))) = "23.4 g NaBr}$