Ironsulfide has 36.5% S by mass. Ironsulfide is heated in oxygen, which makes SO2(g) + ironoxide containing 27.6% O by mass. a.Write a balnced equation for reaction. b.If 1.0 kg of the iron sulfide reacts w excess O, what’s theoretical yield of ironoxide?

The balanced equation is ##”3FeS”+”5O”_2####rarr####”Fe”_3″O”_4″+3SO”_2″##.
The theoretical yield of ##”Fe”_3″O”_4## is 880 g. (Rounded to two due to 1.0 kg.)

A) BALANCED EQUATION
In order to write a balanced chemical equation for this reaction, we must determine the formulas for iron sulfide and iron oxide, as they have more than one formula because iron forms more than one ion.
Determine the formulas for iron sulfide and iron oxide.
Iron Sulfide
S=36.5%
Fe=63.5%
Because 36.5% is 35/100, we can assume a 100-g sample of iron sulfide, and we can say that there are 36.5 g S and 63.5 g Fe. We can convert the masses to moles by dividing the mass of each element by its molar mass (atomic mass on in grams.
Sulfur
##(36.5 “g S”)/((32.065 “g S”)/(“1 mol S”))=((36.5 cancel”g S”)xx(1 “mol S”)/(32.065 cancel”g S”))=1.138 “mol S”##
Iron
##(63.5 “g Fe”)/((55.845 “g Fe”)/(1 “mol S”))=((63.5 cancel”g Fe”)xx(1 “mol Fe”)/(55.845 cancel”g Fe”))=1.137 “mol Fe”##
for ##”Fe”## and ##”S”##
##(1.138 cancel”mol”)/(1.137 cancel”mol”)=1.000##
Iron and sulfur are present in a 1:1 ratio.
The formula for iron sulfide is ##”FeS”##.
Iron Oxide
Again, because 27.6% is 27.6/100, we assume a 100-g sample of iron oxide, and we can say that there are 27.6 g oxygen and 72.4 g Fe. Again, we can convert the masses to moles by dividing the mass of each element by its molar mass (atomic mass on the periodic table in grams.
Oxygen
##(27.6 “g O”)/((15.999 “g O”)/(1 “mol O”))=((27.6 “g O”)xx(1″mol O”)/(15.999 “g O”))=1.725 “mol O”##
Iron
##(72.4 “g Fe”)/((55.845 “g Fe”)/(1″mol Fe”))=((72.4 cancel”g Fe”)xx(1″mol Fe”)/(55.845 cancel”g Fe”))=1.296 “mol O”##
Mole ratios for O and Fe
Divide the moles of each element by the smallest number of moles.
##”O”:####=####(1.725″mol”)/(1.296″mol”)=1.331##
##”Fe”:####=####(1.296″mol”)/(1.296″mol”)=1.000##
Mole ratios must be in whole numbers, so multiply both ratios times ##3##.
##”O”:####=####(1.725″mol”)/(1.296″mol”)=1.331xx3=3.993~~4##
##”Fe”:####=####(1.296″mol”)/(1.296″mol”)=1.000xx3=3##
ratio for ##”Fe:O”####=####3:4##
The chemical formula for iron oxide is ##”Fe”_3″O”_4″##.
Balanced Chemical Equation
##”3FeS”+”5O”_2″####rarr####”3SO”_2″+Fe”_3″O”_4″##
B) THEORETICAL YIELD OF IRON OXIDE
We need the molar masses of ##”FeS”## and ##”Fe”_3″O”_4″## and the mole ratio from the balanced equation.
Molar mass of ##”FeS”=(1xx55.845 “g/mol”)+(1xx32.065 “g/mol”)=87.910 “g/mol”##
Molar mass of ##”Fe”_3″O”_4=(3xx55.945 “g/mol”)+(4xx15.999
“g/mol”)=231.531 “g/mol”##
Mole ratio of iron oxide and iron sulfide is:
##(1 “mol” “Fe”_3″O”_4)/(3 “mol FeS”)##
Calculate Theoretical Yield of Iron Oxide
Convert 1 kg FeS to 1000 g FeS. Then convert the mass of FeS to moles using its molar mass. Multiply moles FeS times the mole ratio between iron oxide and iron sulfide to get moles iron oxide. Multiply moles of iron oxide times its molar mass.
##1000 cancel”g FeS”xx(1cancel”mol FeS”)/(87.91 cancel”g FeS”)xx(1cancel”mol Fe”_3″O”_4)/(3cancel”mol FeS”)xx(231.531 “g Fe”_3″O”_4)/(1cancel”mol Fe”_3″O”4)=880 “g Fe”_3″O”_4″##
(answer rounded to two significant figures because of 1.0 kg)

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