# How many grams of O_2 are needed to react with 84.5 g of NH_3?

May 24, 2018

$\text{198 g O"_2}$ are needed to react with $\text{84.5 g NH"_3}$.

#### Explanation:

Balanced equation

$\text{4NH"_3("g") + "5O"_2("g")}$$\rightarrow$$\text{4NO(g) + 6H"_2"O(g)}$

There are three steps required to answer this question.

$\textcolor{red}{1}$. Convert mass ${\text{NH}}_{3}$ to mol $\text{NH"_3}$ by dividing the given mass by its molar mass $\left(\text{17.031 g/mol}\right)$. Do this by multiplying by the reciprocal of the molar mass.

$\textcolor{b l u e}{2}$. Determine ${\text{mol O}}_{2}$ by multiplying $\text{mol NH"_3}$ by the mol ratio between ${\text{NH}}_{3}$ and ${\text{O}}_{2}$ in the balanced equation, with ${\text{mol O}}_{2}$ in the numerator.

$\textcolor{g r e e n}{3}$. Determine mass ${\text{O}}_{2}$ by multiplying $\text{mol O"_2}$ by its molar mass $\left(\text{31.998 g/mol}\right)$.

color(red)(84.5)color(black)cancel(color(red)("g NH"_3))xx(color(red)1color(black)cancel(color(red)("mol NH"_3)))/(color(red)17.031color(black)cancel(color(red)("g NH"_3)))xx(color(blue)5color(black)cancel(color(blue)("mol O"_2)))/(color(blue)4color(black)cancel(color(blue)("mol NH"_3)))xx(color(green)31.998color(black)cancel(color(green)("g O"_2)))/(color(green)1color(black)cancel(color(green)("mol O"_2)))=color(green)("198 g O"_2" (rounded to three significant figures)

$\text{198 g O"_2}$ are needed to react with $\text{84.5 g NH"_3}$.