(a) Al³⁺ (aq) + CO₃²⁻ (aq) → Al₂(CO₃)₃ (s)
(b) molar concentration of AlCl₃ = 0.108 M
(c) mass of Al₂(CO₃)₃ = 0.379 g
Explanation:
(a) net ionic equation:
Al³⁺ (aq) + CO₃²⁻ (aq) → Al₂(CO₃)₃ (s)
where:
(aq) - aqueous
(s) - solid
(b) chemical equation
2 AlCl₃ + 3 Na₂CO₃ → Al₂(CO₃)₃ + 6 NaCl
molar concentration = number of moles / volume
number of moles = molar concentration × volume
number of moles of Na₂CO₃ = 0.137 M × 35.5 mL = 4.86 mmoles
Talking in account the chemical reaction we devise the following reasoning:
if 2 mmoles of AlCl₃ react with 3 mmoles of Na₂CO₃
then X mmoles of AlCl₃ react with 4.86 mmoles of Na₂CO₃
X = (2 × 4.86) / 3 = 3.24 mmoles of AlCl₃
molar concentration = number of moles / volume
molar concentration of AlCl₃ = 3.24 mmoles / 30 mL
molar concentration of AlCl₃ = 0.108 M
(c) Talking in account the chemical reaction we devise the following reasoning:
if 2 mmoles of AlCl₃ produces with 1 mmole of Al₂(CO₃)₃
then 3.24 mmoles of AlCl₃ react with Y mmoles of Al₂(CO₃)₃
Y = (3.24 × 1) / 2 = 1.62 mmoles of Al₂(CO₃)₃
number of moles = mass / molecular weight
mass = number of moles × molecular weight
mass of Al₂(CO₃)₃ = 1.62 × 234 = 379 mg = 0.379 g
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molar concentration
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