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Given the following chemical equation, if 58 grams of aluminum reacts with excess chlorine gas and was found to produce 203 grams of aluminum chloride, what is the percent yield for this reaction?2Al + 3 Cl2 --> 2AlCl3
35%
24%
71%
68%

Respuesta :

1st find mol of Al = 58/27 = 2.148 mol
There is a 1:1 ratio between AlCl3 and Al so 2.148 mol should be produced
take 2.148 * 133.5 = 286.75 g should be produced

% yield = 203/286.75 = .707 = 71% or C
Oseni

Answer:

70.82%

Explanation:

Recall that: mole = mass/molar mass

From the equation of the reaction; [tex]2Al + 3 Cl_2 --> 2AlCl_3[/tex]

2 moles of aluminium gives 2 moles of AlCl3

58/26.98 moles of aluminium will give x/133.34 moles of AlCl3

x = 2.1497 x 133.34

  = 286.64 g

58 grams of aluminium is supposed to yield 286.64 grams of aluminium chloride. If 203 grams of AlCl3 was produced,

Percentage yield = 203/286.64 x 100%

                                  = 70.82%

The percentage of the reaction is 70.82%

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