How many grams of silver chloride will be precipitated by adding sufficient silver nitrate to react with 1500. mL of 0.400 M barium chloride solution?

2 $AgNO_3(aq) + BaCl_2(aq) \rightarrow 2 AgCl(s) + Ba(NO_3)_2(aq)$

Question

How many grams of silver chloride will be precipitated by adding sufficient silver nitrate to react with 1500. mL of 0.400 M barium chloride solution?

2 [latex]AgNO_3(aq) + BaCl_2(aq) \rightarrow 2 AgCl(s) + Ba(NO_3)_2(aq) [/latex]

Accepted Answer

READ Knowns M = 0.400 M [latex]BaCl_2[/latex]
V = 1500. mL [latex]BaCl_2[/latex]

PLAN Solve as a stoichiometry problem.
Determine the mol [latex]BaCl_2[/latex] in 1500. mL of 0.400 M solution.

mol = [latex]M imes V = (1.500 \cancel{L}) (\frac{0.400 mol BaCl_2}{\cancel{L}} )=0.600 mol BaCl_2[/latex]

Solution map: moles [latex]BaCl_2 ightarrow moles AgCl ightarrow grams [/latex]

CALCULATE [latex](0.600 \cancel{mol BaCl_2}) (\frac{2 \cancel{mol AgCl}}{1 \cancel{mol BaCl_2}} ) (\frac{143.4 g AgCl}{\cancel{mol AgCl}} )=[/latex]172 g AgCl

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