Question

A student has a mixture of salt (NaCl) and sugar (C12H22O11). To determine the percent composition, the student measures out 5.84 g of the mixture and dissolves the sample in 100.0 mL of water. The student knows aqueous NaCl will react with aqueous silver nitrate (AgNO3) to form solid silver chloride (AgCl). The student determines the sugar will not react with silver nitrate. Write a balanced net ionic equation for the reaction of NaCl with AgNO3. How many mL of 1.0 M AgNO3 will be required to precipitate 5.84 g of AgCl

Answer 1

1. Net ionic equation:

• Cl⁻(aq) + Ag⁺(aq) → AgCl(s)

2. Volume of 1.0M AgNO₃

• 41ml

Step-by-step explanation:

1. Net ionic equation for the reaction of NaCl with AgNO₃.

i) Molecular equation:

It is important to show the phases:

• (aq) for ions in aqueous solution
• (s) for solid compounds or elements
• (g) for gaseous compounds or elements

• NaCl(aq) + AgNO₃(aq) → AgCl(s) + NaNO₃(aq)

ii) Dissociation reactions:

Determine the ions formed:

• NaCl(aq) → Na⁺(aq) + Cl⁻(aq)

• AgNO₃(aq) → Ag⁺(aq) + NO₃⁻(aq)

• NaNO₃(aq) → Na⁺(aq) + NO₃⁻(aq)

iii) Total ionic equation:

Substitute the aqueous compounds with the ions determined above:

• Na⁺(aq) + Cl⁻(aq) + Ag⁺(aq) + NO₃⁻(aq) → AgCl(s) + Na⁺(aq) + NO₃⁻(aq)

iv) Net ionic equation

Remove the spectator ions:

• Cl⁻(aq) + Ag⁺(aq) → AgCl(s) ← answer

2. How many mL of 1.0 M AgNO₃ will be required to precipitate 5.84 g of AgCl

i) Determine the number of moles of AgNO₃

The reaction is 1 to 1: 1 mole of AgNO₃ produces 1 mol of AgCl

The number of moles of AgCl is determined using the molar mass:

• number of moles = mass in grams / molar mass

• molar mass of AgCl = 143.32g/mol

• number of moles = 5.84g / (143.32g/mol) = 0.040748 mol

ii) Determine the volume of AgNO₃

• molarity = number of moles of solute / volume of solution in liters

• 1.0M = 0.040748mol / V

• V = 0.040748mol / (1.0M) = 0.040748 liter

• V = 0.040748liter × 1,000ml / liter = 40.748 ml

Round to two significant figures: 41ml ← answer