Question

If 60grams of $NO$ is reacted with sufficient amounts of ${O_2}$ to form $N{O_2}$, which is removed during the reaction, how many grams of $N{O_2}$ can be produced?
(Molar mass: $NO$ = 30g/mol, $N{O_2}$ = 46g/mol)
A.46g
B.60g
C.92g
D.120g
E.180g

Answer 1

We can calculate the grams of $N{O_2}$ using the molar mass of $N{O_2}$, $NO$ and the grams of $NO$. The mass of $NO$ in grams is converted into a number of moles of $NO$ using the molar mass of $NO$. The number of moles of reactant is converted into the number of moles of the product with help of mole-mole conversion factor. The number of moles of $N{O_2}$ is converted into the number of grams $N{O_2}$ with the help of the molar mass.

Complete step by step answer:
Given data contains,
Mass of $NO$ reacted is 60grams.
Molar mass of $NO$ is 30g/mol.
Molar mass of $N{O_2}$ is 46g/mol.
Nitrogen monoxide reacts with oxygen to give the product nitrogen dioxide. We can write this equation as,
$NO + {O_2} \to N{O_2}$
So, one mole of nitrogen monoxide gives one mole of nitrogen dioxide.
We can first calculate the moles of the reactant $NO$. We have to calculate the moles of $NO$ from the grams of $NO$ using the molar mass.
Moles of $NO$ $= 60gNO \times \dfrac{{1mol}}{{30gNO}} = 2molNO$
The moles of $NO$ is $2mol$.
So, let us now convert the number of moles of $NO$ to the number of moles of $N{O_2}$ using the mole-mole conversion factor.
Moles of N{O_2}$$$ = 2mol\,NO \times \dfrac{{2molN{O_2}}}{{2molNO}} = 2molN{O_2}$$ The moles of N{O_2}$is$2mol$. So, we can convert the number of moles of$N{O_2}$into the grams of$N{O_2}$using the molar mass. Grams of$N{O_2}$$ = 2molN{O_2} \times \dfrac{{46gN{O_2}}}{{1molN{O_2}}} = 92gN{O_2}$The grams of$N{O_2}$is$92g$.

Therefore, the option (C) is correct.

Note:
We can also solve this problem directly without the conversion of moles of reactant $NO$and product $N{O_2}$
Given data contains,
Mass of $NO$ reacted is 60grams.
Molar mass of $NO$ is 30g/mol.
Molar mass of $N{O_2}$ is 46g/mol.
Nitrogen monoxide reacts with oxygen to give the product nitrogen dioxide. We can write this equation as,
$NO + {O_2} \to N{O_2}$
So, one mole of nitrogen monoxide gives one mole of nitrogen dioxide.
Therefore, 30g of nitrogen monoxide gives 46g of nitrogen dioxide.
Thus, 60g of nitrogen monoxide would produce,
Grams of nitrogen dioxide$= \dfrac{{60gNO}}{{30g/molNO}} \times 46g/molN{O_2}$
Grams of nitrogen dioxide$= 92gN{O_2}$.
The grams of nitrogen dioxide is $92g$.