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?

Md Ariful Islam

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The first thing you have to do is determine the formula for iron sulfide by using the known of sulfur in the compound. Let's assume that the formula for iron sulfide looks like this

$Fe_xS_y$

The of sulfur in this formula would be

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

This means that you have

$y * 32.07 * 100 = x * 55.85 * 36.5 + y * 32.07 * 36.5$

$3207*y = 20.38.525*x + 1170.555*y$

$2036.445*y=2038.555*x => x/y = 2036.445/2038.555 = 0.9990 ~=1$

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 $FeS$.

The chemical equation looks like this

$FeS + O_2 -> Fe_aO_b + SO_2$

Now do the same for the iron oxide.

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

This will get you

$1600*b = 1541.46*a + 441.6*b$

$1158.4*b = 1541.46*a => a/b = 1158.4/1541.46 = 0.751 ~=3/4$

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 means that your balanced chemical equation will look like this

$color(red)(3)FeS_((s)) + 5O_(2(g)) -> Fe_3O_(4(s)) + 3SO_(2(g))$

Notice the $color(red)(3):1$ 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.

Since oxygen is in excess, all the moles of iron sulfide will take part in the reaction. Use the compound's molar mass to determine how many moles you have in your 1.0-kg sample

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

This means that the reaction will produce

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

The theoretical yield of the reaction will thus be

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

Rounded to two , the number of sig figs you gave for the mass of iron sulfide, the answer will be

$m_(Fe_3O_4) = color(green)("880 g = 0.88 kg")$

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