Question

You have a solid 7.00 gram mixture of sodium nitrate and silver nitrate. You add distilled water to dissolve the solids. Now you have aqueous solutions of sodium nitrate and silver nitrate. Next you add excess sodium chloride which results in a precipitate forming. You collect and dry the precipitate that forms and it has a mass of 2.54 grams. Write a balanced net ionic equation for the reaction that occurred. Determine the percent silver nitrate in the original mixture by mass assuming 95.9% actual yield.

Answer 1

Answer:

Net ionic equation: Ag⁺(aq) + Cl⁻(aq) → AgCl(s)

44.9% as AgNO₃

Explanation:

When sodium nitrate, NaNO₃ and silver nitrate, AgNO₃ are dissolved in water, the Na⁺, NO₃⁻ and Ag⁺ ions are formed.

Then, the addition of NaCl (Na⁺ and Cl⁻) produce AgCl⁻ as precipitate. The net ionic equation is:

Ag⁺(aq) + Cl⁻(aq) → AgCl(s)

If 2.54g of AgCl are formed and represents the 95.9% of yield. The real amount of AgCl is:

2.54g AgCl * (100% / 95.9%) = 2.65g AgCl.

In moles (Molar mass AgCl = 143.32g/mol):

2.65g AgCl * (1mol / 143.32g) = 0.0185 moles of AgCl = Moles of AgNO₃

Because all Ag comes from AgNO₃

Thus, the original mass of silver nitrate and its precentage is (Molar mass AgNO₃ = 169.87g/mol):

0.0185 moles AgNO₃ * (169.87g / mol) = 3.14g of AgNO₃

Percentage:

3.14g AgNO₃ / 7.00g * 100 =  

44.9% as AgNO₃