A compound containing #"36.5%"# of sulfur and #"63.5%"# of iron reacts with a compound containing #"27.6%"# oxygen and #"724%"# of iron. What is the balanced equation and the theoretical yield of iron oxide?

Answer 1

The theoretical yield of the iron oxide will be 880 g.

The first step is to use the known percent composition of sulfur in the compound to find the formula for iron sulfide. Assuming that the formula for iron sulfide looks like this

#Fe_xS_y#

In this formula, the percentage of sulfur would be

#(y * 32.07cancel("g/mol"))/((x * 55.85 + y * 32.07)cancel("g/mol")) * 100 = 36.5#

This indicates that you've

#y * 32.07 * 100 = x * 55.85 * 36.5 + y * 32.07 * 36.5#
#3207*y = 20.38.525*x + 1170.555*y#
#2036.445*y=2038.555*x => x/y = 2036.445/2038.555 = 0.9990 ~=1#
The ratio between iron ans dulfur in rion sulfide is #1:1#. This means that the chemical formula of the compound will have to be #FeS#.

This is how the chemical equation appears.

#FeS + O_2 -> Fe_aO_b + SO_2#

Proceed similarly with the iron oxide now.

#(b * 16.0cancel("g/mol"))/((a * 55.85 + b * 16.0)cancel("g/mol")) * 100 = 27.6#

You'll get this

#1600*b = 1541.46*a + 441.6*b#
#1158.4*b = 1541.46*a => a/b = 1158.4/1541.46 = 0.751 ~=3/4#
If you take #b# to be #4/3*a#, your oxide will look like this
#Fe_aO_((4/3)*a) = Fe_1O_(4/3)# #-># multiply by 3 to get #Fe_3O_4#.

This implies that the form of your balanced chemical equation will be as follows.

#color(red)(3)FeS_((s)) + 5O_(2(g)) -> Fe_3O_(4(s)) + 3SO_(2(g))#
Notice the #color(red)(3):1# mole ratio that exists between iron sulfide and iron (II, III) oxide.. This tells you that, regardless of how many moles of iron sulfide react, your reaction will produce 5/3 fewer moles of the iron oxide.

Because there is too much oxygen present, every mole of iron sulfide will participate in the reaction. To find out how many moles of the compound are in your 1.0-kg sample, use its molar mass.

#1.0cancel("kg") * (1000cancel("g"))/(1cancel("kg")) * ("1 mole "FeS)/(87.91cancel("g")) = "11.38 moles"# #FeS#

This indicates that the response will result in

#11.38cancel("moles"FeS) * ("1 mole "Fe_3O_4)/(color(red)(3)cancel("moles"FeS)) = "3.793 moles"# #Fe_3O_4#

Thus, the reaction's theoretical yield will be

#3.793cancel("moles"Fe_3O_4) * "231.53 g"/(1cancel("mole"Fe_3O_4)) = "878.19 g"# #Fe_3O_4#

The answer, rounded to two sig figs, will be the number of sig figs you provided for the mass of iron sulfide.

#m_(Fe_3O_4) = color(green)("880 g = 0.88 kg")#
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Answer 2

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 significant figures due to 1.0 kg.)

A) EQUAL PERCENTAGE

We need to find the formulas for iron sulfide and iron oxide because iron can form multiple ions, and we need to do this in order to write a balanced chemical equation for this reaction.

Find the iron oxide and iron sulfide formulas.

Sulfur Dioxide

Fe=63.5% S=36.5%

By dividing the mass of each element by its molar mass (atomic mass on the periodic table in grams), we can convert the masses to moles. Since 36.5% is 35/100, we can assume a 100-g sample of iron sulfide and state that there are 36.5 g S and 63.5 g Fe.

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"#
Mole ratios for #"Fe"# and #"S"#
#(1.138 cancel"mol")/(1.137 cancel"mol")=1.000#

Sulfur and iron are found in a 1:1 ratio.

The formula for iron sulfide is #"FeS"#.

Oxide of Iron

Once more, we can assume a 100-g sample of iron oxide because 27.6% is 27.6/100. Based on this, we can say that there are 27.6 g of oxygen and 72.4 g of Fe. Once more, we can convert the masses to moles by dividing each element's mass in grams by its molar mass, or atomic mass on the periodic table.

Air

#(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 proportions of Fe and O

Calculate the smallest number of moles by dividing the moles of each element.

#"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#
The mole ratio for #"Fe:O"##=##3:4#
The chemical formula for iron oxide is #"Fe"_3"O"_4"#.

Equilibrated Chemical Formula

#"3FeS"+"5O"_2"##rarr##"3SO"_2"+Fe"_3"O"_4"#

B) IRON OXIDE THEORICAL YIELD

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"#

The mole ratio between iron sulfide and oxide is:

#(1 "mol" "Fe"_3"O"_4)/(3 "mol FeS")#

Determine the Theoretical Iron Oxide Yield

To calculate moles of iron oxide, multiply moles of FeS by the mole ratio of iron oxide to iron sulfide, or 1 kg FeS to 1000 g FeS. Next, use the molar mass of FeS to convert its mass to moles. Finally, multiply the moles of iron oxide by their 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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Answer 3

The balanced equation for the reaction between sulfur-iron compound and oxygen-iron compound is:

4FeS + 7O2 → 2Fe2O3 + 4SO2

To find the theoretical yield of iron oxide (Fe2O3), we need to calculate the moles of each reactant, determine the limiting reactant, and then use stoichiometry to find the moles of Fe2O3 produced. Finally, we convert moles of Fe2O3 to grams.

Given the percentages:

  • The molar mass of FeS is approximately 87.91 g/mol.
  • The molar mass of Fe2O3 is approximately 159.69 g/mol.

Now, we can calculate the moles of each reactant:

  • Moles of FeS = (36.5 g / 100 g/mol) / 87.91 g/mol
  • Moles of O2 = (27.6 g / 100 g/mol) / 32 g/mol
  • Moles of FeS = (724 g / 100 g/mol) / 159.69 g/mol

Next, we determine the limiting reactant, which is the one that produces the least amount of product. After finding the limiting reactant, we use stoichiometry to find the moles of Fe2O3 produced.

Finally, we convert moles of Fe2O3 to grams using its molar mass.

This process will give us the theoretical yield of iron oxide.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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