Question

A student was given a 2.850-g sample of a mixture of potassium nitrate and potassium bromide and was asked to find the percentage of each compound in the mixture. She dissolved the sample and added a solution that contained an excess of silver nitrate, AgNO3. The silver ion precipitated all of the bromide ion in the mixture as AgBr. It was filtered, dried, and weighed. Its mass was 1.740 g. What was the percentage of each compound in the mixture

Answer 1

See explanation

Step-by-step explanation:

The reaction occurs as follows;

KBr(aq) + AgNO3(aq) ----> AgBr(s) + KNO3(aq)

Number of moles of AgBr formed = mass /molar mass =1.740 g/187.77 g/mol = 0.0093 moles

From the reaction equation;

1 mole of KBr yields 1 mole of AgBr

Hence the number of moles of KBr reacted = 0.0093 moles

Mass of KBr present = 0.0093 moles × 119g/mol = 1.11 g

Mass of KNO3 = 2.850 g - 1.11 g = 1.74 g

Percentage of KBr = 1.11 g/2.850 g × 100 = 38.9%

Percentage of KNO3 = 1.74 g/2.850 g × 100 = 61.1%