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julsineya [31]
3 years ago
12

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
Chemistry
1 answer:
seraphim [82]3 years ago
8 0

<u>Answer:</u>

<u>For A:</u> Aluminium is the limiting reagent.

<u>For B:</u> Oxygen gas is the limiting reagent.

<u>For C:</u> Aluminium is the limiting reagent.

<u>For D:</u> Aluminium is the limiting reagent.

<u>Explanation:</u>

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)

  • <u>For A:</u>  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.

  • <u>For B:</u>  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.

  • <u>For C:</u>  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.

  • <u>For D:</u> 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.

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