Question

A sample of gallium bromide, GaBr3, weighing 0.165 g was dissolved in water and treated with silver nitrate, AgNO3, resulting in the precipitation of 0.299 g AgBr. Use these data to compute the %Ga (by mass) GaBr3.

Answer 1

Answer:

%Ga by mass in sample of $GaBr_{3}$ is 23.0%

Explanation:

Molar mass of AgBr = 187.77 g/mol

So, 0.299 g of AgBr = $\frac{0.299}{187.77}$ moles of AgBr = $\frac{0.299}{187.77}$ moles of Br = 0.00159 moles of Br

Br in AgBr comes from $GaBr_{3}$

So, there are 0.00159 moles of Br in 0.165 g of $GaBr_{3}$

Molar mass of Br = 79.904 g/mol

So, mass of Br in 0.165 g of $GaBr_{3}$ = $(0.00159\times 79.904)g$ = 0.127 g

So, mass of Ga in sample of $GaBr_{3}$ = (0.165-0.127) g = 0.038 g

So, %Ga by mass in sample of $GaBr_{3}$ = [(mass of Ga)/(mass of $GaBr_{3}$)]$\times 100$%

                                                              = $\frac{0.038}{0.165}\times 100$ %

                                                              = 23.0%