Question

For the reaction shown, find the limiting reactant for each of the following initial amounts of reactants. 4Al(s)+3O2(g)→2Al2O3(s) A) 1 molAl, 1 mol O2 B) 4 molAl, 2.6 mol O2 C) 16 molAl, 13 mol O2 D) 7.4 molAl, 6.5 mol O2

Answer 1

Answer:

For A: Aluminium is the limiting reagent.

For B: Oxygen gas is the limiting reagent.

For C: Aluminium is the limiting reagent.

For D: Aluminium is the limiting reagent.

Explanation:

Limiting reagent is defined as the reagent which is present in less amount and it limits the formation of products.

Excess reagent is defined as the reagent which is present in large amount.

For the given chemical reaction:

$4Al(s)+3O_2(g)\rightarrow 2Al_2O_3(s)$

• For A:  1 mol Al, 1 mol $O_2$

By stoichiometry of the reaction:

4 moles of aluminium reacts with 3 moles of oxygen gas

So, 1 mole of aluminum will react with = $\frac{3}{4}\times 1=0.75moles$ of oxygen gas.

As, the given amount of oxygen gas is more than the required amount. Thus, it is considered as an excess reagent.

Hence, aluminium is the limiting reagent.

• For B:  4 mol Al, 2.6 mol $O_2$

By stoichiometry of the reaction:

3 moles of oxygen gas reacts with 4 moles of aluminium

So, 2.6 moles of oxygen gas will react with = $\frac{4}{3}\times 2.6=3.458moles$ of aluminium.

As, the given amount of aluminium is more than the required amount. Thus, it is considered as an excess reagent.

Hence, oxygen gas is the limiting reagent.

• For C:  16 mol Al, 13 mol $O_2$

By stoichiometry of the reaction:

4 moles of aluminium reacts with 3 moles of oxygen gas

So, 16 mole of aluminum will react with = $\frac{3}{4}\times 16=12moles$ of oxygen gas.

As, the given amount of oxygen gas is more than the required amount. Thus, it is considered as an excess reagent.

Hence, aluminium is the limiting reagent.

• For D: 7.4 mol Al, 6.5 mol $O_2$

By stoichiometry of the reaction:

4 moles of aluminium reacts with 3 moles of oxygen gas

So, 7.4 mole of aluminum will react with = $\frac{3}{4}\times 7.4=5.55moles$ of oxygen gas.

As, the given amount of oxygen gas is more than the required amount. Thus, it is considered as an excess reagent.

Hence, aluminium is the limiting reagent.

Answer 2

Limiting reactant are defined as the reactant that is involved primarily in the reaction and formation of product depends on this reactant. the limiting reactant for the given equation are as follows:

• A: Aluminium is the limiting reagent.
• B: Oxygen gas is the limiting reagent.
• C: Aluminium is the limiting reagent.
• D: Aluminium is the limiting reagent.

Limiting reagent are those that limits the formation of products, whereas the reagent present in the excess amount are known as excess reagent.

For A:

For the given chemical reactions, limiting reagents are:

• $\text {4 Al}+ \text{3 O}_2\rightarrow\;\text{2Al}_2\text O_3$

By applying the stereochemistry concept:

4 moles of Al reacts with 3 moles of oxygen.

• 1 mole of Al will react $\dfrac {3}{4}\times 1&= 0.75$ moles of oxygen.

Hence, the given amount of oxygen is more than the required amount, it is an excess reagent and aluminum is limiting reagent.

• For B:

Similarly, in the chemical reaction, where 4 moles of aluminum react with 2.6 moles of oxygen, such that:

3 moles of oxygen reacts with 4 moles of aluminum.

• so, 2.6 moles will react with $\dfrac {3}{4}\times 2.6&= 3.458$ moles of aluminum.

The given amount of aluminium is more than the required amount, aluminum is considered as an excess reagent and oxygen gas is the limiting reagent.

• For C:

Also, 16 moles of aluminum reacts with 13 moles of oxygen.

4 moles of aluminum reacts with 3 moles of oxygen, such that:

• 16 moles of aluminum will react with $\dfrac {3}{4}\times 16&= 12$ moles of oxygen.

Thus, oxygen is excess reagent and aluminum is limiting reagent.

• For D:

7.4 mol Aluminum reacts with 6.5 moles of oxygen, such that:

• 7.4 moles of aluminum will react with $\dfrac {3}{4}\times 7.4 &= 5.55$ moles of oxygen gas.

Therefore, the aluminum is limiting reagent and oxygen is excess reagent.

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