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

The balanced net ionic equation for the reaction between AgNO3 and NaCl is Ag+(aq) + Cl-(aq) → AgCl(s). To determine the percent silver nitrate in the mixture, the mass of the precipitate AgCl is used to back-calculate the initial mass of AgNO3, taking into account the actual yield of 95.9%. The percent composition of AgNO3 is then found to be 41.1%.

Step-by-step explanation:

The net ionic equation represents the species that actually change during the reaction. For the reaction between aqueous solutions of silver nitrate (AgNO₃) and sodium chloride (NaCl), we can write the net ionic equation as follows:

Ag+(aq) + Cl-(aq) → AgCl(s)

Now, to determine the percent silver nitrate in the original mixture by mass (assuming a 95.9% actual yield), we use the mass of the precipitate, AgCl, obtained. From stoichiometry, we know that 1 mole of AgNO₃ will produce 1 mole of AgCl, and since the molar mass of AgNO₃ is 169.87 g/mol and the molar mass of AgCl is 143.32 g/mol, we can calculate the mass of AgNO₃ that would have been present if 100% yield were obtained:

2.54 g AgCl x (1 mol AgCl / 143.32 g AgCl) x (169.87 g AgNO₃ / 1 mol AgNO₃) = mass of AgNO₃ at 100% yield

Mass of AgNO₃ at 100% yield = 2.54 g / 143.32 * 169.87 = 3.00 g

However, considering a 95.9% yield:

Actual mass of AgNO₃ = Theoretical mass of AgNO₃ x Actual yield % = 3.00 g x 95.9% = 2.877 g

Finally, to find the percent composition of silver nitrate in the original mixture:

Percent AgNO₃ = (Mass of AgNO₃ / Total mass of mixture) x 100 = (2.877 g / 7.00 g) x 100

Percent AgNO₃ = 41.1%

Therefore, the percent silver nitrate in the original mixture by mass is 41.1%.

Answer 2

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

44.9% as AgNO₃

Step-by-step 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