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3 years ago

Under the right conditions aluminum will react with chlorine to produce aluminium chloride.

Chemistry

1 answer:

Andreas93 [3]3 years ago

5 0

Answer:

The Limiting reagent is Aluminum. Its not in excess. So the reaction ends once its exhausted. So all Aluminum will be used up

From the question ....

157g of Al

Find the number if reacting Moles...

M=mass/molar mass

M= 157/27

=5.81moles of Aluminum is reacting.

Now looking at the equation of reaction(Always make sure its balanced)...

2moles of Al react to produce 2 Moles of AlCl3

Their Mole ratio is equal

Since their Mole ratio is equal 2:2.... That means... That 5.81moles of AlCl3 would be produced because the mole of Al is 5.81

So We now know the Moles of AlCL3 produced

getting the Mass won't be difficult.

Again

Mole=Mass/Mm

Mass = Mole x Mm(Molar Mass)

Mass of AlCl3 = 5.81 x (27 + 35.5x3)

Note: The arithmetic done in the parenthesis is the molar mass of AlCl3

Mass of AlCl3 = 5.81 x 133.5

=775.63g of AlCl3.

Have a great Day!!!!

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You are given a solid that is a mixture of Na2SO4 and K2SO4. A 0.205-g sample of the mixture is dissolved in water. An excess of

Svetach [21]

Answer:

Mass of SO₄⁻² = 0.123 g.

Mass percentage of SO₄⁻² = 41.2%

Mass of Na₂SO₄ = 0.0773 g

Mass of K₂SO₄ = 0.1277 g

Explanation:

Here we have

We place Na₂SO₄ = X and

K₂SO₄ = Y

Therefore

X +Y = 0.205 .........(1)

Therefore since the BaSO₄ is formed from BaCl₂, Na₂SO₄ and K₂SO₄ we have

Amount of BaSO₄ from Na₂SO₄ is therefore;

$X\times\frac{Molar \, Mass \, of \, BaSO_4}{Molar \, Mass \, of \, Na_2SO_4}$

Amount of BaSO₄ from K₂SO₄ is;

$Y\times\frac{Molar \, Mass \, of \, BaSO_4}{Molar \, Mass \, of \, K_2SO_4}$

Molar mass of

BaSO₄ = 233.38 g/mol

Na₂SO₄ = 142.04 g/mol

K₂SO₄ = 174.259 g/mol

$X\times\frac{Molar \, Mass \, of \, BaSO_4}{Molar \, Mass \, of \, Na_2SO_4}$ = $X\times\frac{233.38 }{142.04}$ = 1.643·X

$Y\times\frac{Molar \, Mass \, of \, BaSO_4}{Molar \, Mass \, of \, K_2SO_4}$ = $Y\times\frac{233.38 }{174.259 }$ = 1.339·Y

Therefore, we have

1.643·X + 1.339·Y = 0.298 g.....(2)

Solving equations (1) and (2) gives

The mass of SO₄⁻² in the sample is given by

Mass of sample = 0.298

Molar mass of BaSO₄ = 233.38 g/mol

Mass of Ba = 137.327 g/mol

∴ Mass of SO₄ = 233.38 g - 137.327 g = 96.05 g

Mass fraction of SO₄⁻² in BaSO₄ = 96.05 g/233.38 g = 0.412

Mass of SO₄⁻² in the sample is 0.412×0.298 = 0.123 g.

Percentage mass of SO₄⁻² = 41.2%

Solving equations (1) and (2) gives

X = 0.0773 g and Y = 0.1277 g.

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</span>

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