A 25.0 mL sample of 0.200 M AgNO3(aq) was allowed to react with an excess of NaCI(aq). The AgCl precipitate from the reaction was carefully dried and weighed. How many grams of precipitate should be obtained? AgNO3(aq) + NACI(aq) → AgCI(s) + NANO3(aq) О 3.13g O 0.717 g O 0.432 g O 1.70g O 1.08 g
A 25.0 mL sample of 0.200 M AgNO3(aq) was allowed to react with an excess of NaCI(aq). The AgCl precipitate from the reaction was carefully dried and weighed. How many grams of precipitate should be obtained? AgNO3(aq) + NACI(aq) → AgCI(s) + NANO3(aq) О 3.13g O 0.717 g O 0.432 g O 1.70g O 1.08 g
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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A 25.0 mL sample of 0.200 M AgNO3(aq) was allowed to react with an excess of NaCI(aq). The AgCI precipitate from the reaction was
carefully dried and weighed. How many grams of precipitate should be obtained?
AgNO3(aq) + NACI(aq) → A£CI(s) + NaNO3(ag)
O 3.13g
O 0.717 g
O 0.432 g
O 1.70 g
O 1.08 g
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Expert Solution

Step 1
Given,
A 25.0 mL sample of 0.100 M AgNO3(aq) was allowed to react with an excess of NaCl(aq).
Volume = 25.0 mL = 0.025 L
Molality = 0.200 M
Molarity = number of moles/volume in L
Number of moles = Molarity x volume in L
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