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How many molecules of aluminum oxide can be produced if 2.68 1023 molecules of aluminum reacts

with 1.88 1023 molecules of oxygen gas?
How much of which reactant will
remain?

1 Answer

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To determine the number of molecules of aluminum oxide produced, we first need to determine the limiting reactant. The limiting reactant is the one that is completely consumed in the reaction and determines the maximum amount of product that can be formed.

Let's calculate the number of moles for each reactant:

Number of moles of aluminum = number of molecules / Avogadro's number

Number of moles of aluminum = 2.68 × 10^23 / 6.022 × 10^23 = 0.445 moles

Number of moles of oxygen gas = number of molecules / Avogadro's number

Number of moles of oxygen gas = 1.88 × 10^23 / 6.022 × 10^23 = 0.312 moles

The balanced equation for the reaction between aluminum and oxygen gas to form aluminum oxide is:

4 Al + 3 O2 → 2 Al2O3

According to the stoichiometry of the balanced equation, 4 moles of aluminum react with 3 moles of oxygen gas to produce 2 moles of aluminum oxide.

To find the limiting reactant, we compare the moles of aluminum and oxygen gas with the stoichiometric ratio:

Aluminum: Oxygen gas = 4:3

Since the ratio of moles of aluminum to moles of oxygen gas is 0.445 moles / 0.312 moles ≈ 1.43, we can see that there are fewer moles of oxygen gas compared to the stoichiometric ratio. Therefore, oxygen gas is the limiting reactant.

The stoichiometric ratio tells us that for every 3 moles of oxygen gas, we produce 2 moles of aluminum oxide. Thus, the number of moles of aluminum oxide produced is:

Number of moles of aluminum oxide = (2/3) × number of moles of oxygen gas

Number of moles of aluminum oxide = (2/3) × 0.312 moles = 0.208 moles

To find the number of molecules of aluminum oxide, we multiply the number of moles by Avogadro's number:

Number of molecules of aluminum oxide = number of moles × Avogadro's number

Number of molecules of aluminum oxide = 0.208 moles × 6.022 × 10^23 = 1.25 × 10^23 molecules

Therefore, 1.25 × 10^23 molecules of aluminum oxide can be produced.

To calculate how much of each reactant remains, we need to determine the amount of the limiting reactant consumed.

The number of moles of oxygen gas consumed is equal to the number of moles of oxygen gas initially present minus the number of moles of oxygen gas used in the reaction:

Number of moles of oxygen gas consumed = initial moles of oxygen gas - moles of oxygen gas used

Number of moles of oxygen gas consumed = 0.312 moles - 0.312 moles = 0 moles

Since all the oxygen gas is consumed, there is none remaining.

To find the amount of aluminum remaining, we use the stoichiometry of the balanced equation:

4 moles of aluminum react with 3 moles of oxygen gas

Since the moles of oxygen gas consumed is 0, it means that all 4 moles of aluminum react and none remain.

In conclusion:

Number of molecules of aluminum oxide produced: 1.25 × 10^23 molecules

Amount of oxygen gas remaining: 0 moles

Amount of aluminum remaining: 0 moles

User Danzel
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