Answer : The mass of iron(III)sulfide is, 5.4288 g
Solution : Given,
Mass of iron, Fe = 3 g
Mass of sulfur, = 2.5 g
Molar mass of Fe = 56 g/mole
Molar mass of = 256 g/mole
Molar mass of iron(III)sulfide, = 208 g/mole
The balanced chemical reaction is,
First we have to calculate the moles of iron and sulfur.
From the balanced reaction, we conclude that
16 moles of Fe react with 3 moles of
0.054 moles of Fe react with moles of
Therefore, the excess reagent in this reaction is, Fe and limiting reagent is,
Now we have to calculate the moles of FeS.
As, 3 moles of gives 8 moles of
So, 0.0098 moles of gives moles of
The moles of = 0.0261 moles
Now we have to calculate the mass of .
Mass of = Moles of × Molar mass of
Mass of = 0.0261 g × 208 g/mole = 5.4288 g
Therefore, the mass of iron(III)sulfide is, 5.4288 g
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