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IgorLugansk [536]
3 years ago
12

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.
Chemistry
1 answer:
melamori03 [73]3 years ago
8 0

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%

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