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soldi70 [24.7K]
2 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]2 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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I hope this helps you.

8 0
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Is potassium fluoride a covalent bond or ionic bond
Ivahew [28]

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Hope this helped!

7 0
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What is the luster of aluminum
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Answer:

Luster of aluminum is Silver-White.

Explanation:

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3 0
2 years ago
What is the mass of 3.62X10^24 molecules of methanol
Bad White [126]
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7 0
3 years ago
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Determine the number of grams in each of the quantities<br><br> 1.39.0 x 1024 molecules Cl2
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Mass of Cl₂ : 164.01 g

<h3>Further explanation</h3>

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Can be formulated :

N = n x No

N = number of particles

n = mol

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mol Cl₂ :

\tt n=\dfrac{N}{No}\\\\n=\dfrac{1.39.10^{24}}{6.02.10^{23}}\\\\n=2.31

mass Cl₂(MW=71 g/mol) :

\tt mass=mol\times MW\\\\mass=2.31\times 71=164.01

8 0
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