How many O2 are needed to react with 8.95 moles of Al?

The answer to your question is 6.713 moles of O₂ or 107.4 grams

Explanation:

Data

moles and grams of O₂

moles of Al = 8.95

Balanced chemical reaction

                4Al  +  3 O₂   ⇒    2Al₂O₃

Process

1.- Use proportions and cross multiplication to find the answer. Use the coefficients of the balanced reaction.

               4 moles of Al 3 moles of O₂

                8.95 moles   x

                x = (8.95 x 3) / 4

                x = 26.85 / 4

               x = 6.713 moles of O₂

2.- Calculate the grams of O₂

                 16 g of O₂ 1 mol

                   x                6.713 moles

                  x = (6.713 x 16) / 1

                 x = 107.4 grams

We need 6.71 moles of O2 (or 214.7 grams of O2) to react with 8.95 moles of Al

Explanation:

Step 1: Data given

Number of moles of Al = 8.95 moles

Step 2: The balanced equation

4Al + 3O2 → 2Al2O3

Step 3: Calculate moles number of moles O2 needed to react with Al

For 4 moles Al we need 3 moles O2 to produce 2 moles Al2O3

For 8.95 moles Al we need 3/4 * 8.95 = 6.71 moles O2

There will 8.95/2 = 4.48 moles Al2O3 be produced

We need 6.71 moles of O2 to react with 8.95 moles of Al

Mass O2 = moles O2 * molar mass O2

Mass O2 = 6.71 moles * 32.0 g/mol

Mass O2 =214.7 grams O2

We need 214.7 grams of O2 to react with 8.95 moles Al

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