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1 year ago

A 2.80 g sample of Al reacts with 4.15 g sample of Cl2 according to the equation shown below.

Chemistry

1 answer:

solong [7]1 year ago

4 0

Answer:

Mass = 5.33 g

Explanation:

Given data:

Mass of Al = 2.80 g

Mass of Cl₂ = 4.15 g

Theoretical yield of AlCl₃ = ?

Solution:

Chemical equation:

2Al + 3Cl₂ → 2AlCl₃

Number of moles of Al:

Number of moles = mass/molar mass

Number of moles = 2.80 g/ 27 g/mol

Number of moles = 0.10 mol

Number of moles of Cl₂:

Number of moles = mass/molar mass

Number of moles = 4.15 g/71 g/mol

Number of moles = 0.06 mol

Now we will compare the moles of AlCl₃ with Al and Cl₂.

Cl₂ : AlCl₃

3 : 2

0.06 : 2/3×0.06 = 0.04

Al : AlCl₃

2 : 2

0.10 : 0.10

Number of moles of AlCl₃ produced by chlorine are less so it will be limiting reactant.

Mass of AlCl₃:Theoretical yield

Mass = number of moles ×molar mass

Mass = 0.04 mol × 133.34 g/mol

Mass = 5.33 g

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2 years ago

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Dennis_Churaev [7]

Answer:

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2 years ago

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