Question

How many grams of iron (3)sulfide are produced when 131 grams of iron react with excess sulfur?

How many grams of Al will completely replace 25.5 grams of Cu when Aluminum is mixed with aqueous copper (2) sulfate?

Answer 1

We want the molar mass (gfm) of Fe and FeS.

Fe = 55.8 g/mol and FeS = 87.8 g/mol (55.8 +32.0).

Explanation:  

We want  ratio between Fe:FeS 8:8. This comes from the coefficients of the equal chemical equation.

Now we set up conversion factors following the roadmap from above.

Unit we want in the numerator, unit to cancel in the denominator.

Multiply the numerators, and divide the denominators.

$(131.0)gFe \times 1molFe/(55.8)gFe \times 8molFeS/8molFe \times 87.8gFes/1molFes$

$243.82 g \ of Fe_2S_3$

The outcome is 243.82 g of ferrous sulphite

By the above given formula we can find any amount of Fes produced by any amount of  ferrous react with sulphite.

Answer 2

Answer:

1) 243.3 g of Fe III sulphide

2) 7.2g of Al

Explanation:

1)

2Fe(s) + 3S(s) -----> Fe2S3(s)

Since iron is the limiting reagent

From the reaction equation:

112g of Fe produces 208g of Iron III sulphide

131 g of Fe will produce 131×208/112= 243.3 g of Fe III sulphide

2)

2Al(s) + 3CuSO4(aq) -------> Al2(SO4)3(aq) + 3Cu(s)

From the reaction equation:

54 g of Al displaced 192 g of Cu

xg of Al will displace 25.5 g of Cu

x= 25.5×54/192= 7.2g of Al