Question

4 g of Mg is reacted with 8 g of O2. Mole of MgO produced in the reaction is

Answer 1

Answer:1/6 mole of MgO is produced

Explanation:the clear cut solution is provided in the attachment

Answer 2

Answer : The moles of MgO produced in the reaction is 0.125 moles.

Solution : Given,

Mass of Mg = 4 g

Mass of $O_2$ = 8 g

Molar mass of Mg = 24 g/mole

Molar mass of $O_2$ = 32 g/mole

Molar mass of MgO = 40 g/mole

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

$\text{ Moles of }Mg=\frac{\text{ Mass of }Mg}{\text{ Molar mass of }Mg}=\frac{4g}{24g/mole}=0.167moles$

$\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=\frac{8g}{32g/mole}=0.25moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$2Mg(s)+O_2(g)\rightarrow 2MgO(s)$

From the balanced reaction we conclude that

As, 2 mole of $Mg$ react with 1 mole of $O_2$

So, 0.25 moles of $Mg$ react with $\frac{0.25}{2}=0.125$ moles of $O_2$

From this we conclude that, $O_2$ is an excess reagent because the given moles are greater than the required moles and $Mg$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $MgO$

From the reaction, we conclude that

As, 2 mole of $Mg$ react with 2 mole of $MgO$

So, 0.125 moles of $Mg$ react with 0.125 moles of $MgO$

Therefore, the moles of MgO produced in the reaction is 0.125 moles.