Question

A 0.8838-g sample of an ionic compound containing bromide ions and an unknown metal cation is dissolved in water and treated with an excess of AgNO3. If 1.573 g of a AgBr precipitate forms, what is the percent by mass of Br in the original compound?

Answer 1

Answer:

%$Br=75.7$%

Explanation:

Hello,

From the AgBr precipitate, one could compute one can compute the bromine grams as follows:

$1.573gAgBr*\frac{80gBr}{188gAgBr}=0.669gBr$

As long as there was an excess of silver nitrate, one knows that into the 1.573 g of AgBr, 0.669 g correspond to the bromine that was initially contained into the 0.8838-g sample, thus, the percent is computed as follows:

%$Br=\frac{0.669gBr}{0.8838g}*100$%

%$Br=75.7$%

Best regards.

Answer 2

Wow this is really hard dude