Question

. Both KBr and NaBr are soluble ionic compounds and fully dissociate in aqueous solution. A 21.545-g mixture of KBr and NaBr is dissolved in water, then a solution of AgNO3 is added so that all of the bromine present is converted to solid AgBr. The AgBr product is dried and found to have a mass of 35.593 g. What mass of KBr was present in the original mixture?

Answer 1

Step-by-step explanation:

Let the molar mass of KBr is
`M_(1)` and NaBr is
`M_(2)`.

In 21.545 g, the moles of KBr and NaBr are
`n_(1)` and
`n_(2)`.

Therefore,
`M_(1) * n_(1) + M_(2) * n_(2)` = 21.545 g

`n_(2) = (21.545 g - M_(1) * n_(1))/(M_(2))`

Mass of AgBr is 35.593 g.

`(M_(Ag) + M_(Br))(n_(1) + n_(2))` = 35.593 g

`(M_(Ag) + M_(Br))n_(1) + (M_(Ag) + M_(Br))n_(2)` = 35.593 g

`M_(AgBr) + (M_(Ag) + M_(Br))(21.545 g - M_(1) * n_(1))/(M_(2))` = 35.593 g

`n_(1) = (35.593 g * M_(2) - M_(AgBr) * 21.545 g)/(M_(AgBr) (M_(2) - M_(1)))`

Now, putting the given values into the above formula as follows.

`n_(1) = (35.593 g * M_(2) - M_(AgBr) * 21.545 g)/(M_(AgBr) (M_(2) - M_(1)))`

`n_(1) = (35.593 g * 102.894 g/mol - 187.7722 g/mol * 21.545 g)/(187.7722 g/mol (109.894 g/mol - 119.002 g/mol))`

=
`(3662.306 - 4045.552)/(-1710.229) mol`

=
`(383.246)/(1710.229) mol`

= 0.224 mol

Hence, mass of KBr will be calculated as follows.

`0.224 mol (119.002 g)/(1 mol)`

= 26.656 g

Thus, we can conclude that mass of KBr present in the original mixture is 26.656 g.