what is the theoretical yield of nabr when 2.36 moles of febr3 reacts? 2febr3 + 3na2s → fe2s3 + 6nabr ? mol…

what is the theoretical yield of nabr when 2.36 moles of febr3 reacts? 2febr3 + 3na2s → fe2s3 + 6nabr ? mol nabr round your answer to the hundredths place.

what is the theoretical yield of nabr when 2.36 moles of febr3 reacts? 2febr3 + 3na2s → fe2s3 + 6nabr ? mol nabr round your answer to the hundredths place.

Answer

Explanation:

Step1: Identify mole - ratio

From the balanced equation $2FeBr_3 + 3Na_2S\rightarrow Fe_2S_3+6NaBr$, the mole - ratio of $FeBr_3$ to $NaBr$ is $\frac{n_{NaBr}}{n_{FeBr_3}}=\frac{6}{2} = 3$.

Step2: Calculate moles of $NaBr$

Given $n_{FeBr_3}=2.36$ moles. Using the mole - ratio, $n_{NaBr}=3\times n_{FeBr_3}$. $n_{NaBr}=3\times2.36$ moles. $n_{NaBr}=7.08$ moles.

Answer:

$7.08$