A5.000 g mixture contains strontium nitrate and potassium bromide. excess lead(ii) nitrate solution is added to precipitate out 0.7822 grams of lead(ii) bromide. what is the percent by mass of potassium bromide in the mixture?

The mass percent of potassium bromide in the mixture is 9.996%

Explanation:

To calculate the number of moles, we use the equation:

    .....(1)

For lead (II) bromide:

Given mass of lead (II) bromide = 0.7822 g

Molar mass of lead (II) bromide = 367 g/mol

Putting values in equation 1, we get:

The chemical equation for the reaction of lead (II) nitrate and potassium bromide follows:

By Stoichiometry of the reaction:

1 mole of lead (II) bromide is produced from 2 moles of potassium bromide

So, 0.0021 moles of lead (II) bromide will be produced from = of potassium bromide

Now, calculating the mass of potassium bromide by using equation 1, we get:

Molar mass of KBr = 119 g/mol

Moles of KBr = 0.0042 moles

Putting values in equation 1, we get:

To calculate the percentage composition of KBr in the mixture, we use the equation:

Mass of mixture = 5.000 g

Mass of KBr = 0.4998 g

Putting values in above equation, we get:

Hence, the percent by mass of KBr in the mixture is 9.996 %

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