Question

how many moles fe2s3 would be produced from the complete reaction of 449 g febr3? 2febr3 + 3na2s → fe2s3 + 6nabr

Explanation:

Step1: Calculate the molar mass of \(FeBr_3\)

The molar mass of \(Fe\) is \(55.85\ g/mol\), and the molar mass of \(Br\) is \(79.90\ g/mol\).
For \(FeBr_3\), \(M = 55.85+(3\times79.90)=55.85 + 239.7=295.55\ g/mol\)

Step2: Calculate the number of moles of \(FeBr_3\)

Using the formula \(n=\frac{m}{M}\), where \(m = 449\ g\) and \(M = 295.55\ g/mol\)
\(n_{FeBr_3}=\frac{449\ g}{295.55\ g/mol}\approx1.52\ mol\)

Step3: Use the stoichiometry of the reaction

From the balanced equation \(2FeBr_3+3Na_2S
ightarrow Fe_2S_3 + 6NaBr\), the mole ratio of \(FeBr_3\) to \(Fe_2S_3\) is \(2:1\)
Let \(n_{Fe_2S_3}\) be the moles of \(Fe_2S_3\). Then \(n_{Fe_2S_3}=\frac{1}{2}n_{FeBr_3}\)
Substitute \(n_{FeBr_3}=1.52\ mol\)
\(n_{Fe_2S_3}=\frac{1}{2}\times1.52\ mol = 0.76\ mol\)

Answer:

\(0.76\)