Question

Magnesium oxide can be made by heating magnesium metal in the presence of oxygen. The balanced equation for the reaction is: 2Mg(s)+O2(g)→2MgO(s) When 10.1 g of Mg are allowed to react with 10.5 g of O2, 10.3 g of MgO are collected. Dteremine the limiting reactants for the reaction

Answer 1

Answer: The limiting reactants for the reaction is, Mg

Explanation : Given,

Mass of
`Mg` = 10.1 g

Mass of
`O_2` = 10.5 g

Molar mass of
`Mg` = 24 g/mol

Molar mass of
`O-2` = 32 g/mol

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

`\text{Moles of }Mg=\frac{\text{Given mass }Mg}{\text{Molar mass }Mg}`

`\text{Moles of }Mg=(10.1g)/(24g/mol)=0.421mol`

and,

`\text{Moles of }O_2=\frac{\text{Given mass }O_2}{\text{Molar mass }O_2}`

`\text{Moles of }O_2=(10.5g)/(32g/mol)=0.328mol`

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation 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, 0.421 moles of
`Mg` react with
`(0.421)/(2)=0.2105` 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.

Hence, the limiting reactants for the reaction is, Mg