A 5.95-g sample of AgNO3 is reacted with BaCl2 according to the equation to give 4.07 g of AgCl. What is the percent yield of AgCl?

The % yield of AgCl is 81.1 %

Explanation:

Step 1: Data given

Mass of AgNO3 = 5.95 grams

Molar mass AgNO3 = 169.87 g/mol

Molar mass BaCl2 = 208.23 g/mol

Mass of AgCl = 4.07 grams

Molar mass of AgCl = 143.32 g/mol

Step 2: The balanced equation

2AgNO3 + BaCl2 → 2AgCl + Ba (NO3) 2

Step 3: Calculate moles AgNO3

Moles AgNO3 = mass AgNO3 / molar mass AgNO3

Moles AgNO3 = 5.95 grams / 169.87 g/mol

Moles AgNO3 = 0.0350 moles

Step 4: Calculate moles AgCl

For 2 moles AgNO3 we need 1 mol BaCl2 to produce 2 moles AgCl and 1 mol Ba (NO3) 2

For 0.0350 moles AgNO3 we'll have 0.0350 moles AgCl

Step 5: Calculate mass AgCl

Mass AgCl = moles AgCl * moles AgCl

Mass AgCl = 0.0350 moles * 143.32 g/mol

Mass AgCl = 5.016 grams

Step 6: Calculate % yield AgCl

% AgCl = (4.07 grams / 5.016 grams) * 100 %

% AgCl = 81.1 %

The % yield of AgCl is 81.1 %