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

Explanation:

Step1: Calculate molar - mass of FeBr₃

The molar mass of Fe (iron) is approximately 55.85 g/mol, Br (bromine) is approximately 79.90 g/mol. For FeBr₃, (M = 55.85+3\times79.90=55.85 + 239.7=295.55) g/mol.

Step2: Calculate moles of FeBr₃

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

Step3: Use mole - ratio from the balanced equation

From the balanced equation (2FeBr_{3}+3Na_{2}S\rightarrow Fe_{2}S_{3}+6NaBr), the mole - ratio of (FeBr_{3}) to (Fe_{2}S_{3}) is (2:1). Let the moles of (Fe_{2}S_{3}) be (n_{Fe_2S_3}). Then (n_{Fe_2S_3}=\frac{1}{2}n_{FeBr_3}).

Step4: Calculate moles of Fe₂S₃

Substitute (n_{FeBr_3}=1.52) mol into the equation. (n_{Fe_2S_3}=\frac{1}{2}\times1.52 = 0.76) mol.

Answer:

0.76 mol