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MA_775_DIABLO [31]
3 years ago
14

Ill give the brainliest answer to whoever helps me with this equation

Chemistry
1 answer:
vampirchik [111]3 years ago
6 0

Answer: The percent yield for the NaBr is, 86.7 %

Explanation : Given,

Moles of FeBr_3 = 2.36 mol

Moles of NaBr = 6.14 mol

First we have to calculate the moles of NaBr

The balanced chemical equation is:

2FeBr_3+3Na_2S\rightarrow Fe_2S_3+6NaBr

From the reaction, we conclude that

As, 2 moles of FeBr_3 react to give 6 moles of NaBr

So, 2.36 moles of FeBr_3 react to give \frac{6}{2}\times 2.36=7.08 mole of NaBr

Now we have to calculate the percent yield for the NaBr.

\text{Percent yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield = 6.14 moles

Theoretical yield = 7.08 moles

Now put all the given values in this formula, we get:

\text{Percent yield}=\frac{6.14mol}{7.08mol}\times 100=86.7\%

Therefore, the percent yield for the NaBr is, 86.7 %

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<h3>General Formulas and Concepts:</h3>

<u>Math</u>

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Order of Operations: BPEMDAS

  1. Brackets
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<u>Step 2: Identify Conversions</u>

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<u>Step 3: Convert</u>

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