Question

Aluminum ions react with carbonate ions to form an insoluble compound, aluminum carbonate. (a) Write the net ionic equation for this reaction. (b) What is the molarity of a solution of aluminum chloride if 30.0 mL is required to react with 35.5 mL of 0.137 M sodium carbonate? (c) How many grams of aluminum carbonate are formed in (b)

Answer 1

(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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