In each reaction, calculate the moles of product (in parentheses) when 3.00 moles of each reactant is used.

A: 4Li+O2 = 2Li2O (Li2O)
B: Fe2O3+3H2 = 2Fe+3H2O (Fe)
C: Al2S3+6H2O = 2Al(OH)3+3H2S (H2S)

Please show work, Thanks

a. if you have 3 moles of O2, 3 moles of Li, you are limited by the Li. You should get then the mole ratio of product to Li (2/4) times the moles of Li, or 2/4*3 moles Li2O

b. You will be limited by three moles of H2. You get then 2Fe, and 3H2o

c. you will be limited by the H2o, so you get 1 aluminum hydroxide, and 1.5 H2S moles

Sure! To calculate the moles of product, you need to use the balanced chemical equation and the stoichiometry of the reaction. The coefficients in the balanced equation represent the mole ratios between the reactants and products.

Let's solve each reaction one by one:

Reaction A: 4Li + O2 → 2Li2O

Given that 3.00 moles of each reactant is used, we start by converting moles of Li to moles of Li2O.

Using the stoichiometry from the balanced equation, we can see that the mole ratio between Li and Li2O is 4:2. This means that for every 4 moles of Li, we will get 2 moles of Li2O.

So, to calculate the moles of Li2O, we can set up a proportion:

(3.00 moles Li) / (4 moles Li) = (x moles Li2O) / (2 moles Li2O)

Cross-multiplying and solving for x, we get:

x = (3.00 moles Li * 2 moles Li2O) / 4 moles Li
= 1.50 moles Li2O

Therefore, when 3.00 moles of each reactant is used in Reaction A, we will produce 1.50 moles of Li2O.

Now, let's move to Reaction B:

Reaction B: Fe2O3 + 3H2 → 2Fe + 3H2O

Using the same approach, we start by converting the moles of Fe2O3 to moles of Fe.

The stoichiometry tells us that the mole ratio between Fe2O3 and Fe is 1:2. So, for every 1 mole of Fe2O3, we get 2 moles of Fe.

Setting up the proportion:

(3.00 moles Fe2O3) / (1 mole Fe2O3) = (x moles Fe) / (2 moles Fe)

Solving for x:

x = (3.00 moles Fe2O3 * 2 moles Fe) / 1 mole Fe2O3
= 6.00 moles Fe

Hence, with 3.00 moles of each reactant in Reaction B, we will produce 6.00 moles of Fe.

Finally, let's solve Reaction C:

Reaction C: Al2S3 + 6H2O → 2Al(OH)3 + 3H2S

To find the moles of H2S, we convert the moles of Al2S3.

The stoichiometry shows that the mole ratio between Al2S3 and H2S is 1:3. So, for every 1 mole of Al2S3, we get 3 moles of H2S.

Setting up the proportion:

(3.00 moles Al2S3) / (1 mole Al2S3) = (x moles H2S) / (3 moles H2S)

Solving for x:

x = (3.00 moles Al2S3 * 3 moles H2S) / 1 mole Al2S3
= 9.00 moles H2S

Thus, when 3.00 moles of each reactant is used in Reaction C, the result is 9.00 moles of H2S.

To summarize:
- In Reaction A, you will get 1.50 moles of Li2O.
- In Reaction B, you will obtain 6.00 moles of Fe.
- In Reaction C, you will produce 9.00 moles of H2S.