Respuesta :
The balanced equation for the reaction is as follows;
2Fe + 3S ---> Fe₂S₃
molar ratio of Fe to Fe₂S₃ is 2:1
mass of Fe₂S₃ to be produced is - 96 g
therefore number of moles of Fe₂S₃ to be produced is - 96 g / 208 g/mol
number of Fe₂S₃ moles = 0.46 mol
according to the molar ratio
when 2 mol of Fe reacts with 3 mol of sulfur then 1 mol of Fe₂S₃ is produced
that for 1 mol of Fe₂S₃ to be produced - 2 mol of Fe should react
therefore for 0.46 mol of Fe₂S₃ to be produced - 2 x 0.46 = 0.92 mol of Fe is required
mass of Fe required - 0.92 mol x 56 g/mol = 51.5 g
mass of Fe required is - 51.5 g
The mass of iron that is needed to chemically react with sulfur in order to produce 96 grams of iron (III) sulfide is 51.688 grams.
Given the following data:
- Mass of iron (III) sulfide = 96 grams
Scientific data:
- Molar mass of iron = 56 g/mol.
- Molar mass of iron (III) sulfide = 208 g/mol.
To determine the mass of iron that is needed to chemically react with sulfur in order to produce 96 grams of iron (III) sulfide:
First of all, we would write the properly balanced chemical equation for this chemical reaction:
[tex]2 Fe + 3 S \rightarrow Fe_2S_3[/tex]
Next, we would the number of moles of iron (III) sulfide produced:
[tex]Number\;of\;moles = \frac{Mass}{molar\;mass}\\\\Number\;of\;moles = \frac{96}{208 }[/tex]
Number of moles = 0.4615 moles.
By stoichiometry:
2 mole of iron = 0.4615 moles of iron (III) sulfide
1 mole of iron = [tex]0.4615 \times = 2 = 0.923\;moles[/tex]
For mass of iron:
[tex]Mass = Number\;of\;moles \times molar\;mass\\\\Mass = 0.923 \times 56[/tex]
Mass of iron = 51.688 grams
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