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amid [387]
4 years ago
8

Find the mass of AlCl3 that is produced when 25.0 grams of Al2O3 react with excess HCl according to the following equation.

Chemistry
1 answer:
kondaur [170]4 years ago
8 0

Answer:

Mass = 65.4 g

Explanation:

Given data:

Mass of Al₂O₃ = 25 g

Mass of AlCl₃ = ?

Solution:

Chemical equation:

Al₂O₃ + 6HCl  →   2AlCl₃ + 3H₂O

Number of moles of Al₂O₃:

Number of moles = mass/ molar mass

Number of moles = 25 g/  101.96 g/mol

Number of moles = 0.2452 mol

Now we will compare the moles of Al₂O₃  and AlCl₃.

                          Al₂O₃         :            AlCl₃

                              1             :               2

                             0.2452    :          2×0.2452 = 0.4904 mol

Mass of AlCl₃:

Mass = number of moles × molar mass

Mass = 0.4904 mol × 133.34 g/mol

Mass = 65.4 g

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The new volume will be 1379 mL.

Explanation:

As per Boyle's law, the product of initial volume and initial pressure of any gas molecule is equal to the product of final volume and final pressure of those molecules.

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P_{1} V_{1} = P_{2} V_{2}

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Initiation

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where the big black dot indicates unpaired electrons attached to the atom.

Propagation

\text{CH}_3\text{Cl}+ \text{Cl}\bullet \to \bullet\text{CH}_2\text{Cl} + \text{HCl}

\bullet\text{CH}_2\text{Cl} + \text{Cl}_2 \to \text{CH}_2\text{Cl}_2 + \text{Cl}\bullet

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Termination

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