Question

Consider the balanced reactionbelow:2FeBr3 + 3Na2S → Fe₂S3 +6NaBrHow many moles of iron(III) sulfide,Fe2S3, would be produced from thecomplete reaction of 449 g iron(III)romide

Answer 1

Answer:

0.7596moles

Explanations:

Given the balanced chemical reactions

$2FeBr_3+3Na_2S\rightarrow Fe_2S_3+6NaBr$

Given the following parameters

Mass of FeBr3 = 449grams

Determine the mole of FeBr3

$\begin{gathered} mole\text{ of FeBr}_3=\frac{mass}{molar\text{ }mass} \\ mole\text{ of FeBr}_3=\frac{449}{295.56} \\ mole\text{ of FeBr}_3=1.519moles \end{gathered}$

According to stoichiometry, 2moles of FeBr3 produces 1mole of Fe2S3, the moles of Fe2S3 required is given as:

$\begin{gathered} mole\text{ of Fe}_2S_3=\frac{1mole\text{ of Fe}_2S_3}{2moles\text{ of FeBr}_3}\times1.519moles\text{ of FeBr}_3 \\ mole\text{ of Fe}_2S_3=0.7596moles \end{gathered}$

Hence the mole of Fe2S3 that will be produced is 0.7596moles