# Question 3fb7d

Feb 18, 2017

You have no way of knowing.

#### Explanation:

The information provided by the problem can only get you the volume of carbon dioxide produced by the reaction. In order to determine the mass of carbon dioxide, you must know the pressure and temperature at which the reaction takes place.

So, the balanced chemical equation that describes your reaction looks like this

$\textcolor{b l u e}{2} {\text{CO"_ ((aq)) + "O"_ (2(g)) -> color(darkorange)(2)"CO}}_{2 \left(g\right)}$

Notice that the for every $\textcolor{b l u e}{2}$ moles of carbon monoxide and $1$ mole of oxygen gas consumed by the reaction, you get $\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{2}$ moles of carbon dioxide.

When all the three gases are kept under the same conditions for pressure and temperature, this mole ratio becomes equivalent to a volume ratio.

You can thus say that when $\textcolor{b l u e}{2}$ liters of carbon monoxide react with $1$ liter of oxygen gas, $\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{2}$ liters of carbon dioxide are produced.

Notice that this is exactly how many liters of carbon monoxide and oxygen gas you have, so you know that the reaction will produce $2$ liters of carbon dioxide.

In order to be able to find the mass of the sample, you must know the pressure and temperature at which the reaction takes place.

Let's say, for example, that the reaction takes place at a pressure $P$ $\text{atm}$ and a temperature $T$ $\text{K}$. You can use the ideal gas law equation to find the number of moles of gas, $n$, present in the sample

$P V = n R T \implies n = \frac{P V}{R T}$

Here

• $R$ is the universal gas constant, usually given as $0.0821 \left(\text{atm" * "L")/("mol" * "K}\right)$

In your case, you would have

$n = \left(P \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{atm"))) * 2 color(red)(cancel(color(black)("L"))))/(0.0821(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * T color(red)(cancel(color(black)("K}}}}\right)$

$n = 24.36 \cdot \frac{P}{T} \textcolor{w h i t e}{.} \text{moles}$

To convert this to grams, use the molar mass of carbon dioxide

24.36 * P/T color(red)(cancel(color(black)("moles CO"_2))) * "44.01 g"/(1color(red)(cancel(color(black)("mole CO"_2)))) = (1072 * P/T)" g"#

You can use this general solution for any value of $P$ and $T$, as long as they are expressed in atmospheres and Kelvin, respectively.