Answer to Question #165021 in General Chemistry for Fernando

Question

A mixture consisting of only manganese(II) bromide (MnBr₂) and nickel bromide (NiBr₂) weighs 0.9604 g. When the mixture is dissolved in water and an excess of silver nitrate is added, all the bromide ions associated with the original mixture are precipitated as insoluble silver bromide (AgBr). The mass of the silver bromide is found to be 1.6658 g. Calculate the mass percentages of manganese(II) bromide and nickel bromide in the original mixture.

Mass percent MnBr₂ = %Mass percent NiBr₂ = %

Answer 1

MnBr₂ + AgNO₃ $\to$ Mn(NO₃)₂ + 2AgBr

x/214.7 + . $\to$ . + 2x/214.7

NiBr₂ + AgNO₃ $\to$ Ni(NO₃)₂ + 2AgBr

(0.9604-x)/218 + . $\to$ . + 2(0.9604-x)/218

since mass of AgBr = 1.6658g

so moles = $\frac{1.6658}{187.77}$

$\frac{2x}{214.7}$ + $\frac{2(0.9604-x)}{218}$ = $\frac{1.6658}{187.77}$

x = 0.234

% MnBr₂ = $\frac{0.234}{0.9604}$ ×100 = 24.36%

% NiBr₂ = $\frac{0.9604-0.234}{0.9604}$ ×100 = 75.64%