Verified Answer :

Hagrid

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\(* \frac{mol}{24g} *2 = 8\)
For O2: 88.5g*\(\frac{mol}{32g} *1 = 2.7\)
Since O2 has the smaller moles, this is the Limiting reactant. Then we solve basing on the number of O2 used. 

88.5g\(* \frac{mol}{32g} * \frac{2}{1} * \frac{56g}{mol} = 309.75 g\)

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 \(O_2\).

\(\text{Moles of Mg}=\frac{\text{Mass of Mg}}{\text{Molar mass of Mg}}=\frac{97.2g}{24.3g/mole}=4\text{ moles}\)

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

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 \(\frac{4}{2}=2\) 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.

\(\text{Mass of MgO}=\text{Moles of MgO}\times \text{Molar mass of MgO}\)

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

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

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