How many moles of aluminum oxide are produced according to the reaction below given that you start with 10.0 g of al and 19.0 grams of o2? reaction: 4al + 3o2 → 2al2o3 a) 0.396 b) 0.185 c) 0.741 d) 5.00 e) not enough information

Considering the reaction stoichiometry and the definition of limiting reagent, the correct answer is option B): the total number of moles of Al₂O₃ produced is 0.185 moles.

The balanced reaction is:

6 Al + 3 O₂ → 2 Al₂O₃

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

The molar mass of each compound is:

So, by reaction stoichiometry, the following amount of mass of each compound participate in the reaction:

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 96 grams of O₂ reacts with 108 grams of Al, 19 grams of O₂ react with how much mass of Al?

mass of Al= 21.375 grams

But 21.375 grams of Al are not available, 10 grams are available. Since you have less mass than you need to react with 19 grams of O₂, Al will be the limiting reagent.

Then you can apply the following rule of three to calculated the amount of moles of aluminum oxide produced, using the limitng reagent: if by stoichiometry 108 grams of Al produce 2 moles of Al₂O₃, 10 grams of Al will produce how many moles of Al₂O₃?

amount of moles of Al₂O₃= 0.185 moles

Finally, the correct answer is option B): the total number of moles of Al₂O₃ produced is 0.185 moles.

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The answer to your question is:   x = 0.185 mol of Al2O3

Explanation:

Data

10 g of Al and 19 grams of O2

Reaction:                                4Al       +       3O2       →       2Al2O3

                                            4(27)                3(32)               2 ( 102)

                                             108g                 96g                    204g

Limiting reactant

                                    108g  Al  96g  O2

                                      10 g           x

                                     x = (10 x 96) / 108 = 8.9 g of O2

Then limiting reactant is Al

So

                                  108 g of Al   204 g of Al2O3

                                    10 g             x

                                   x = (10 x 204) / 108 = 18.8 g of Al2O3

                                    102 g of Al2O3 1 mol

                                      18.8                   x

                                  x = 0.185 mol of Al2O3

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