A sample of 523.1 mg of impure KBr is treated with an excess of AgNO3 and 814.5 mg of AgBr is obtained. What is the purity of KBr?

98.68%

Explanation:

Step 1: Write the balanced equation

KBr + AgNO₃ ⇒ AgBr + KNO₃

Step 2: Calculate the moles corresponding to 814.5 mg (0.8145 g) of AgBr

The molar mass of AgBr is 187.77 g/mol.

0.8145 g × 1 mol/187.77 g = 4.338 × 10⁻³ mol

Step 3: Calculate the moles of KBr needed to produce 4.338 × 10⁻³ moles of AgBr

The molar ratio of KBr to AgBr is 1:1. The moles of KBr needed are 1/1 × 4.338 × 10⁻³ mol = 4.338 × 10⁻³ mol.

Step 4: Calculate the pure mass corresponding to 4.338 × 10⁻³ moles of KBr

The molar mass of KBr is 119.00 g/mol.

4.338 × 10⁻³ mol × 119.00 g/mol = 0.5162 g

Step 5: Calculate the purity of KBr

0.5162 g of KBr are in a 0.5231 g-sample. The purity of KBr is:

P = 0.5162 g/0.5231 g × 100% = 98.68%

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