Question

Given 17.2 grams of $MgO$ are produced according to the following equation: $2Mg+{{O}_{2}}\to 2MgO$. How many grams of oxygen ${{O}_{2}}$ must have been consumed?

Answer 1

Hint: The answer is obtained by finding the molar masses of each atom which is then followed by the calculations which include the determination of the oxygen consumed by multiplying the moles of oxygen with its molar mass.

Complete step by step solution:
We are familiar with the concepts that deal with the calculation of molar masses and also several related concepts which help in the determination of the number of moles of the product being produced and also the amount of the reactant consumed etc.
Now, let us determine the amount of oxygen which is being consumed in the reaction in terms of grams.
- To start with, let us determine the molar masses of each atom in the reaction that is $2Mg+{{O}_{2}}\to 2MgO$
- Molar mass of $MgO$is calculated as $24+16=40g/mol$
Also the molar mass of diatomic oxygen will be $16\times 2=32g/mol$
Now, the number of moles of $MgO$ will be $17\times \dfrac{1}{40}=0.426mol$ of $MgO$
Now, multiply the moles of $MgO$ with the mole ratio between ${{O}_{2}}$ and from the balanced chemical equation above and we get the moles of ${{O}_{2}}$ that are consumed.
The calculation is $0.426\times \dfrac{1mol({{O}_{2}})}{2mol(MgO)}=0.21338$
Now, by multiplying this value obtained by the molar mass of oxygen, we get the total mass of oxygen consumed in grams and that will be,
\[17.2g(MgO)\times \dfrac{1mol(MgO)}{40g(MgO)}\times \dfrac{1mol({{O}_{2}})}{2mol(MgO)}\times \dfrac{32g({{O}_{2}})}{1mol({{O}_{2}})}=6.83g\]
Hence, the total amount of oxygen consumed in the reaction will be 6.83g

Note:
Here, it is important to note that the stoichiometry of the reaction plays an important role in the determination of the amount of products or reactants produced or used in a reaction.