Question

How many grams of AgCl are produced from the reation of 135 mL of 0.567 M AgNO3 with an excess of AlCl3

Answer 1

The balanced chemical reaction is

3 AgNO₃ + AlCl₃ --> 3 AgCl + Al(NO₃)₃

3 moles AgNO3 requires 1 mole AlCl3.

moles present in 135 mL of 0.567M AgNO3 = 135 mL x 10^-3 L x 0.567 M = 0.076545 moles

0.076545 moles AgNO3 produces 0.076545 moles of AgCl

mass of AgCl produced = moles of AgCl x molar mass of AgCl

mass of AgCl produced = 0.076545 mol x 143.32 g/mol

mass of AgCl produced = 10.9704294 g = 11.0 grams ----- 3 significant figures

Answer: 11.0 g

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