What mass of precipitate (in g) is formed when 80.0 mL of 0.500 M FeI₃ reacts with excess AgNO₃ in the following chemical reaction? FeI₃(aq) + 3 AgNO₃(aq) → 3 AgI(s) + Fe(NO₃)₃(aq)

Answer 1

80.0 mL of 0.500 M FeI₃

0.0800 liter x 0.500 moles/liter = 0.04 moles

Using the balanced chemical reaction equation, 1 mole of FeI3 results in 3 moles of AgI (which is the solid precipitate).

FeI₃(aq) + 3AgNO₃(aq) → 3 AgI(s) + Fe(NO₃)₃(aq)

0.04 moles x 3 = 0.12 moles of AgI

Calculate the molar mass of AgI by looking up silver and iodine on the periodic table:

Ag = 107.87 g/mol

I = 126.90 g/mol

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AgI = 234.77 g/mol

Using the molar mass, calculate the grams of AgI:

0.12 moles of AgI x 234.77 g/mol = 28.2 grams of AgI