Please help me this is my fourth attempt.

Explanation:

CH4 + 4S ---> CS2 + 2H2S

4) 0.75 mol S × (1 mol CS2/4 mol S) = 0.19 mol CS2

5) 3 mol H2S × (1 mol CH4/2 mol H2S) = 1.5 mol CH4

Fe2O3 + 2Al ---> 2Fe + Al2O3

6) 25 g FeO3 × (1 mol Fe2O3/159.69 g Fe2O3) = 0.16 mol Fe2O3

0.16 mol Fe2O3 × (2 mol Al/1 mol Fe2O3) = 0.32 mol Al

0.32 mol Al × (26.98 g Al/1 mol Al) = 8.6 g Al

7) Given:

45 g Al × (1 mol Al/26.98 g Al) = 1.6 mol Al

85 g Fe2O3 ×(1 mol Fe2O3/159.69 g Fe2O3)

= 0.53 mol Fe2O3

Let's look at how much Fe each reactant will produce:

1.6 mol Al × (2 mol Fe/2 mol Al) = 1.6 mol Fe

0.53 mol Fe2O3 × (2 mol Fe/1 mol Fe2O3) = 1.1 mol Fe

Note that the given amount of Fe2O3 will give us fewer Fe. Therefore, Fe2O3 is the limiting reactant.

8) Al will produce 1.6 mol Fe × (55.845 g Fe/1 mol Fe)

= 89 g Fe

Fe2O3 will produce 1.1 mol Fe × (55.845 Fr/1 mol Fe)

= 61 g Fe

9) Since Fe2O3 is the limiting reactant, the ideal yield of Fe for the reaction is 61 g. If the actual reaction only gave us 25 g Fe. then the percent yield of Fe is

%yield = (25 g Fe/61 g Fe) × 100% = 41%

10) If we only got 25 g Fe, then the amount of Al actually used in the reaction is

25 g Fe × (1 mol Fe/55.845 g Fe) = 0.45 mol Fe

0.45 mol Fe × (2 mol Al/2 mol Fe) = 0.45 mol Al

0.45 mol Al × (26.98 g Al/1 mol Al) = 12 g

Therefore, the leftover amount of Al is

25 g Al - 12 g Al = 13 g Al