Answer:
<u>1. Net ionic equation:</u>
- Cl⁻(aq) + Ag⁺(aq) → AgCl(s)
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<u>2. Volume of 1.0M AgNO₃</u>
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
- V = 0.040748mol / (1.0M) = 0.040748 liter
- V = 0.040748liter × 1,000ml / liter = 40.748 ml
Round to two significant figures: 41ml ← answer
