# In the chemical reaction "C"_6"H"_6"OS" + "8O"_2"rarr"6CO"_2 + "3H"_2"O" + "SO"_2, how many moles of "C"_6"H"_6"OS" are needed to react with "317 g O"_2"?

Jul 12, 2017

$\text{1.24 mol C"_6"H"_6"OS}$ are needed to react with ${\text{317 g O}}_{2}$.

#### Explanation:

Balanced equation:

$\text{C"_6"H"_6"OS" + "8O"_2}$$\rightarrow$${\text{6CO"_2 + "3H"_2"O" + "SO}}_{2}$

First convert the given mass of ${\text{O}}_{2}$ to moles. The molar mass of ${\text{O}}_{2}$ is $\text{31.998 g/mol}$.

317color(red)cancel(color(black)("g O"_2))xx(1"mol O"_2)/(31.998color(red)cancel(color(black)("g O"_2)))="9.91 mol O"_2

Determine the moles $\text{C"_6"H"_6"OS}$ that will react with $\text{317 g O"_2}$ by multiplying moles of ${\text{O}}_{2}$ by the mole ratio between $\text{C"_6"H"_6"OS}$ and ${\text{O}}_{2}$ from the balanced equation so that moles ${\text{O}}_{2}$ are canceled.

9.91color(red)cancel(color(black)("mol O"_2))xx(1"mol C"_6"H"_6"OS")/(8color(red)cancel(color(black)("mol O"_2)))="1.24 mol C"_6"H"_6"OS" (rounded to three sig figs)