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How many moles of lithium metal would need to react with excess oxygen gas to produce 4 moles of lithium oxide in the following reaction?

4Li + O2 → 2Li2O (My answer: C)


1

2

4

8

Respuesta :

4Li + O₂ = 2Li₂O

Li : Li₂O = 4 : 2 = 2 : 1

2 : 1
x : 4

x=4*2/1=8 mol

D) 8

Answer: The correct answer is 8.

Explanation: For the following equation:

[tex]4Li(s)+O_2(g)\rightarrow 2Li_2O(s)[/tex]

As oxygen gas is in excess, so lithium metal is considered as the limiting reagent because it limits the formation of product.

By Stoichiometry,

2 moles of lithium oxide is produced by 4 moles of Lithium

So, 4 moles of lithium oxide will be produced by = [tex]\frac{4}{2}{\times 4[/tex] = 8 moles of lithium metal.

So, the correct answer is 8.

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