# Question a5001

Mar 22, 2016

$\text{736.7 g}$

#### Explanation:

Take a look at the balanced chemical equation for this reaction

${\text{CS"_text(2(l]) + color(red)(2)"O"_text(2(g]) -> "CO"_text(2(g]) + 2"SO}}_{\textrm{2 \left(g\right]}}$

Notice the $1 : \textcolor{red}{3}$ mole ratio that exists between the two reactants. This tells you that the reaction will consume $\textcolor{red}{3}$ moles of oxygen gas for very mole of carbon disulfide that takes part in the reaction.

The problem tells you that you need to determine how many grams of oxygen would be needed to react with $7.674$ moles of carbon disulfide.

Right from the start, you know that the reaction must consume $\textcolor{red}{3}$ moles of oxygen per mole of carbon disulfide, so use this mole ratio to determine how many moles of oxygen would be needed here

7.674 color(red)(cancel(color(black)("moles CS"_2))) * (color(red)(3)color(white)(a)"moles O"_2)/(1color(red)(cancel(color(black)("mole CS"_2)))) = "21.022 moles O"_2#

To find out how many grams of oxygen would contain this many moles, use oxygen gas' molar mass

$21.022 \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{moles O"_2))) * overbrace("32.0 g"/(1color(red)(cancel(color(black)("mole O"_2)))))^(color(purple)("molar mass of O"_2)) = color(green)(|bar(ul(color(white)(a/a)"736.7 g O}}_{2} \textcolor{w h i t e}{\frac{a}{a}} |}}}$