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
