The Notes: `44` g of `C_3H_8` react with `80` g of `O_2` to form `CO_2` and `H_2O.` How many grams of `CO

Friday, February 4, 2011

$\displaystyle{44}$ g of $\displaystyle{C}_{{3}}{H}_{{8}}$ react with $\displaystyle{80}$ g of $\displaystyle{O}_{{2}}$ to form $\displaystyle{C}{O}_{{2}}$ and $\displaystyle{H}_{{2}}{O}.$ How many grams of $\displaystyle{C}{O}_{{2}}$ will be formed?

First, write the balanced equation, this is simple:

$\displaystyle{C}_{{3}}{H}_{{8}}+{5}{O}_{{2}}={3}{C}{O}_{{2}}+{4}{H}_{{2}}{O}.$

Hence each mole of propane $\displaystyle{C}_{{3}}{H}_{{8}}$ requires $\displaystyle{5}$ moles of oxygen $\displaystyle{O}_{{2}}.$

Also we need to compute molar masses: for $\displaystyle{C}_{{3}}{H}_{{8}}$ it is $\displaystyle{3}\cdot{12}+{8}\cdot{1}={44}\frac{{g}}{{{m}{o}{l}}},$ for $\displaystyle{O}_{{2}}$ it is $\displaystyle{2}\cdot{16}={32}\frac{{g}}{{{m}{o}{l}}}$ and for $\displaystyle{C}{O}_{{2}}$ it is $\displaystyle{12}+{2}\cdot{16}={44}\frac{{g}}{{{m}{o}{l}}}.$

Now we can determine what initial substance will react completely: $\displaystyle{44}{g{}}$ of $\displaystyle{C}_{{3}}{H}_{{8}}$ is $\displaystyle{1}$ mole and $\displaystyle{80}{g{}}$ of $\displaystyle{O}_{{2}}$ are $\displaystyle\frac{{80}}{{32}}={2.5}$ moles. $\displaystyle{1}$ mole of $\displaystyle{C}_{{3}}{H}_{{8}}$ requires $\displaystyle{5}$ moles of $\displaystyle{O}_{{2}},$ but there are only $\displaystyle{2.5},$ so all oxygen will be used.

Finally, $\displaystyle{5}$ moles of oxygen make $\displaystyle{3}$ moles of $\displaystyle{C}{O}_{{2}},$ so $\displaystyle{2.5}$ moles make $\displaystyle{1.5}$ moles of $\displaystyle{C}{O}_{{2}},$ and they have the mass of $\displaystyle{1.5}\cdot{44}=$ 66 g. This is the answer.