College Chemistry Homework Help

Question

A 76.0 mL solution of 0.200 M of AgNO₃ reacts with 180.0 mL solution of 0.351M of NaCl to produce silver chloride (AgCl) and sodium nitrate (NaNO₃) according to the balanced equation:

AgNO₃(aq)+NaCl(aq)—>AgCl(s)+NaNO₃(aq)

AgNO₃ is the limiting reactant for the given equation.

Calculate the mass of AgCl produced in the given reaction.

mass of AgCl:

If 0.830 g of AgCl are isolated after carrying out the given reaction, calculate the percent yield of AgCl.

percent yield: %

Answer 1

Limiting reactant problem

Since we are told that AgNO3 is the limiting reactant, we will use the mols of AgNO3 to calculate moles of AgCl that can be formed. We will use the stoichiometry of the balanced equation and the molar mass of AgCl to do this:

moles AgNO3 present = 76.0 ml x 1 L / 1000 ml x 0.200 mol / L = 0.0152 mols AgNO3

moles AgCl formed = 0.0152 mols AgNO3 x 1 mol AgCl / mol AgNO3 = 0.0152 mols AgCl

mass of AgCl formed = 0.0152 mols AgCl x 143 g AgCl / mol = 2.17 g AgCl formed

Percent yield = actual yield / theoretical yield (x100%)

% yield = 0.830 g / 2.17 g (x100%) = 38.2% yield