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The imaginary molecule [tex]AB_3[/tex] has a molar mass of 52.1 g/mol and is produced according to the following balanced equation:

\[ A_2 + 3B_2 \rightarrow 2AB_3 \]

When excess [tex]A_2[/tex] reacts with 1.1 mol [tex]B_2[/tex], 36.3 g of [tex]AB_3[/tex] are produced. What is the percent yield for this reaction?

A. 100%
B. 13%
C. 95%
D. 63%
E. 77%
F. 42%

Answer :

Final answer:

To find the percent yield of the reaction is 95%.

option 3 is correct

Explanation:

To find the percent yield of the reaction, we need to compare the actual yield to the theoretical yield. The molar mass of AB3 is given as 52.1 g/mol and 36.3 g of AB is produced. First, we calculate the number of moles of AB produced by dividing the mass of AB by its molar mass: 36.3 g / 52.1 g/mol = 0.697 mol.

Next, we use the balanced equation to determine the theoretical yield of AB3. According to the equation, 2 moles of AB3 are produced for every 3 moles of B2 that react. Since we have 1.1 mol of B2, the theoretical yield of AB3 is (2/3) * 1.1 mol = 0.733 mol.

Finally, we calculate the percent yield by dividing the actual yield (0.697 mol) by the theoretical yield (0.733 mol) and multiplying by 100: (0.697 mol / 0.733 mol) * 100 = 95%.

Therefore, the percent yield for this reaction is 95%.

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