Question

how many moles febr3 are required to generate 275 g nabr? 2febr3 + 3na2s → fe2s3 + 6nabr ? mol febr3

Explanation:

Step1: Calculate the molar mass of NaBr

The molar mass of Na is \(22.99\space g/mol\) and Br is \(79.90\space g/mol\). So, \(M_{NaBr}=22.99 + 79.90=102.89\space g/mol\).

Step2: Calculate the moles of NaBr

Using the formula \(n=\frac{m}{M}\), where \(m = 275\space g\) and \(M = 102.89\space g/mol\). Then \(n_{NaBr}=\frac{275}{102.89}\approx2.67\space mol\).

Step3: Use the stoichiometry from the balanced equation

From \(2FeBr_3+3Na_2S
ightarrow Fe_2S_3 + 6NaBr\), the mole ratio of \(FeBr_3\) to \(NaBr\) is \(2:6=\frac{1}{3}\). Let \(n_{FeBr_3}\) be the moles of \(FeBr_3\). Then \(n_{FeBr_3}=\frac{1}{3}n_{NaBr}\).
Substitute \(n_{NaBr} = 2.67\space mol\) into the equation: \(n_{FeBr_3}=\frac{1}{3}\times2.67 = 0.89\space mol\).

Answer:

\(0.89\space mol\)