Question

How do you find the mass of $C{{O}_{2}}$ produced in a reaction of $150.0 \ g$ of ${{C}_{6}}{{H}_{12}}$ in sufficient (excess) oxygen if the reaction has a $35.00\%$ yield?

Answer 1

Hint: Mole is represented by the symbol mol and it is generally used to measure the amount of substance in SI units where SI term is related with the System of International units. The term mole is combined with the concept of Avogadro number.

Complete answer:
To know about the mass of $C{{O}_{2}}$ produced in the reaction we firstly want to write down the balanced equation of the combustion reaction of ${{C}_{6}}{{H}_{12}}$ which can be written as follows:
${{C}_{6}}{{H}_{12}}+9{{O}_{2}}\to 6C{{O}_{2}}+6{{H}_{2}}O$
Hence from this equation we consider that 1 mol of ${{C}_{6}}{{H}_{12}}$ contains 6 mol of $C{{O}_{2}}$on reacting with excess of oxygen.

Molar mass of ${{C}_{6}}{{H}_{12}}$ is calculated by $6\times 12+12\times 1=84g/mol$. And the molar mass of $C{{O}_{2}}$ is calculated by $1\times 12+16\times 2=44g/mol$. From this we can say that 84 g of ${{C}_{6}}{{H}_{12}}$ produces $6\times 44$ i.e. 264 g of $C{{O}_{2}}$.
Then 150 g of ${{C}_{6}}{{H}_{12}}$ produces $\dfrac{264}{84}\times 150$ g of $C{{O}_{2}}$.
But the yield given in the question is $35\%$ only so according to this mass of $C{{O}_{2}}$ produced will be calculated as follows:
$\dfrac{264}{84}\times 150\times 35\%$ = 165 g
Hence we can say that $165 \ g$ of $C{{O}_{2}}$ produced in a reaction of $150.0 \ g$ of ${{C}_{6}}{{H}_{12}}$ in sufficient (excess) oxygen if the reaction has a $35.00\%$ yield.

Note: Molar mass of any compound is defined as the mass of all components present in the compound i.e. by adding the individual mass of every component present in substance we can easily calculate the molar mass of that compound.