Question

When 20 g of lithium reacts with 30 g of oxygen, then calculate the mass of Li2O formed. Which is in excess
Li or O2, and by how many grams?​

Answer 1

Answer : The mass of excess reagent $O_2$ is, 7.04 grams

Explanation : Given,

Mass of Li = 20 g

Mass of $O_2$ = 30 g

Molar mass of Li = 6.94 g/mole

Molar mass of $O_2$ = 32 g/mole

First we have to calculate the moles of $Li$ and $O_2$.

$\text{Moles of }Li=\frac{\text{Mass of }Li}{\text{Molar mass of }Li}=\frac{20g}{6.94g/mole}=2.88moles$

$\text{Moles of }O_2=\frac{\text{Mass of }O_2}{\text{Molar mass of }O_2}=\frac{30g}{32g/mole}=0.94moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$4Li+O_2\rightarrow 2Li_2O$

From the balanced reaction we conclude that

As, 4 moles of $Li$ react with 1 mole of $O_2$

So, 2.88 moles of $Li$ react with $\frac{2.88}{4}=0.72$ moles of $O_2$

That means, in the given balanced reaction, $Li$ is a limiting reagent that limits the formation of products and $O_2$ is an excess reagent.

The excess reagent remains $(O_2)$  = 0.94 - 0.72 = 0.22 mole

The excess moles of $O_2$ = 0.22 mole

Now we have to calculate the mass of $O_2$.

$\text{Mass of }O_2=\text{Moles of }O_2\times \text{Molar mass of }O_2$

$\text{Mass of }O_2=(0.22mole)\times (32g/mole)=7.04g$

Therefore, the mass of excess reagent $O_2$ is, 7.04 grams

Answer 2

Answer:

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