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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.

Respuesta :

Answer:

%Ga by mass in sample of [tex]GaBr_{3}[/tex] is 23.0%

Explanation:

Molar mass of AgBr = 187.77 g/mol

So, 0.299 g of AgBr = [tex]\frac{0.299}{187.77}[/tex] moles of AgBr = [tex]\frac{0.299}{187.77}[/tex] moles of Br = 0.00159 moles of Br

Br in AgBr comes from [tex]GaBr_{3}[/tex]

So, there are 0.00159 moles of Br in 0.165 g of [tex]GaBr_{3}[/tex]

Molar mass of Br = 79.904 g/mol

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

So, mass of Ga in sample of [tex]GaBr_{3}[/tex] = (0.165-0.127) g = 0.038 g

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

                                                              = [tex]\frac{0.038}{0.165}\times 100[/tex] %

                                                              = 23.0%

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