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bearhunter [10]
3 years ago
5

If you had excess aluminum, how many moles of aluminum chloride could be produced from 21.0 g of chlorine gas, cl2?

Chemistry
1 answer:
AleksandrR [38]3 years ago
3 0
When aluminum metal is made to contact with chlorine gas (Cl₂), a highly exothermic reaction proceeds. This produces aluminum chloride (AlCl₃) powder. The balanced chemical equation for this reaction is shown below:

2Al(s) + 3Cl₂(g) → 2AlCl₃(s)

Since it was stated that aluminum is in excess, this means that the amount of AlCl₃ produced will only depend on the amount of Cl₂ gas available. The molar mass of Cl₂ is 70.906 g/mol. Using stoichiometry, we have the following equation:

(21.0 g Cl₂/ 70.906 g/mol Cl₂) x 2 mol AlCl₃/ 2 mol Cl₂ = 0.1974 mol AlCl₃

Thus, we have determined that 0.1974 <span>moles of aluminum chloride can be produced from 21.0 g of chlorine gas. </span>
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Solution :

We know that :

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Then, $\Delta T_f = k_f.\frac{w_2}{m_2.w_1}$   ..................(1)

Where,

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Given :

\Delta T_f = 3.14\ ^\circ C,   k_f= 5.12\ ^\circ C/m

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                              =5.12 \ ^\circ C \ kg/mole

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Then putting this values in the equation is (1),

$3.14 = \frac{5.12 \times 24.3}{m_2 \times 0.250}$

$m_2 = \frac{5.12 \times 24.3}{3.14 \times 0.250}$

m_2= 158.49  g/mole

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Answer:

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You can view more details on each measurement unit:

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