Question

How many grams of magnesium oxide can be produced when 97.2 g Mg react with 88.5 g O2?

Answer 1

Answer : The mass of magnesium oxide produced will be, 161.2 g

Solution : Given,

Mass of Mg = 97.2 g

Mass of
`O_2` = 88.5 g

Molar mass of Mg = 24.3 g/mole

Molar mass of
`O_2` = 32 g/mole

Molar mass of MgO = 40.3 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}}=(97.2g)/(24.3g/mole)=4\text{ moles}`

`\text{Moles of }O_2=\frac{\text{Mass of }O_2}{\text{Molar mass of }O_2}=(88.5g)/(32g/mole)=2.7\text{ moles}`

The balanced reaction is,

`2Mg+O_2\rightarrow 2MgO`

As, 2 moles of Mg react with 1 moles of
`O_2`

So, 4 moles of Mg react with
`(4)/(2)=2` moles of
`O_2`

From this we conclude that the
`O_2` is in excess amount and Mg is in limited amount.

Now from the reaction we conclude that

As, 2 moles of Mg react to give 2 moles of MgO

So, 4 moles of Mg react to give 4 moles of MgO

Now we have to calculate the mass of MgO.

`\text{Mass of MgO}=\text{Moles of MgO}* \text{Molar mass of MgO}`

`\text{Mass of MgO}=(4moles)* (40.3g/mole)=161.2g`

Therefore, the mass of magnesium oxide produced will be, 161.2 g

Answer 2

First we establish the chemical reaction equation.
2Mg + O2 = 2MgO.
Then we need to know which of the reactant is limiting. We do that by multiplying the number of moles of reactant to the stoichiomeric coefficient. For Magnesium: 97.2 g
`* (mol)/(24g) *2 = 8`
For O2: 88.5g*
`(mol)/(32g) *1 = 2.7`
Since O2 has the smaller moles, this is the Limiting reactant. Then we solve basing on the number of O2 used.

88.5g
`* (mol)/(32g) * (2)/(1) * (56g)/(mol) = 309.75 g`