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soldi70 [24.7K]
3 years ago
9

Aqueous solutions of aluminum sulfate and barium chloride are mixed, resulting in the precipitate formation of barium sulfate wi

th aqueous aluminum chloride as the other product. (Use the lowest possible coefficients.
Chemistry
1 answer:
denpristay [2]3 years ago
3 0

Answer:

Al2(SO4)3(aq) + 3BaCl2(aq) → 3BaSO4(s) + 2AlCl3(aq)

Explanation:

The question wants you to use the lowest possible coefficients to balance the equation.

According to the question the reaction is as follows ;

Generally, writing the chemical formula requires one to exchange the charge between the cations and anions involved. Example aluminum sulfate has Al3+ and (SO4)2- . cross multiply the charges to get Al2(SO4)3

aluminum sulfate → Al2(SO4)3

barium chloride → BaCl2

barium sulfate → BaSO4

aluminum chloride → AlCl3

Al2(SO4)3(aq) + BaCl2(aq) → BaSO4(s) + AlCl3(aq)

To balance a chemical equation one have to make sure the number of atom of element on the reactant side(left) is equal to the number of atom of elements on the product side(right).

Now, the equation can be be balance with the lowest coefficient as follows;

Al2(SO4)3(aq) + 3BaCl2(aq) → 3BaSO4(s) + 2AlCl3(aq)

The bold numbers is the coefficient use to balance the equation.

The number of atom on the reactant side is equal to the product side. Using aluminium atom as a case study the number of aluminium atom on the reactant side is 2 and the on the product  side it is also  2.

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0.0714 M for the given variables

Explanation:

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Mass of copper(II) acetate: m_{(AcO)_2Cu} = 0.972 g

Volume of the sodium chromate solution: V_{Na_2CrO_4} = 150.0 mL

Molarity of the sodium chromate solution: c_{Na_2CrO_4} = 0.0400 M

Now, when copper(II) acetate reacts with sodium chromate, an insoluble copper(II) chromate is formed:

(CH_3COO)_2Cu (aq) + Na_2CrO_4 (aq)\rightarrow 2 CH_3COONa (aq) + CuCrO_4 (s)

Find moles of each reactant. or copper(II) acetate, divide its mass by the molar mass:

n_{(AcO)_2Cu} = \frac{0.972 g}{181.63 g/mol} = 0.0053515 mol

Moles of the sodium chromate solution would be found by multiplying its volume by molarity:

n_{Na_2CrO_4} = 0.0400 M\cdot 0.1500 L = 0.00600 mol

Find the limiting reactant. Notice that stoichiometry of this reaction is 1 : 1, so we can compare moles directly. Moles of copper(II) acetate are lower than moles of sodium chromate, so copper(II) acetate is our limiting reactant.

Write the net ionic equation for this reaction:

Cu^{2+} (aq) + CrO_4^{2-} (aq)\rightarrow CuCrO_4 (s)

Notice that acetate is the ion spectator. This means it doesn't react, its moles throughout reaction stay the same. We started with:

n_{(AcO)_2Cu} = 0.0053515 mol

According to stoichiometry, 1 unit of copper(II) acetate has 2 units of acetate, so moles of acetate are equal to:

n_{AcO^-} = 2\cdot 0.0053515 mol = 0.010703 mol

The total volume of this solution doesn't change, so dividing moles of acetate by this volume will yield the molarity of acetate:

c_{AcO^-} = \frac{0.010703 mol}{0.1500 L} = 0.0714 M

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