# Thread: Chemistry - Gas Laws question

1. ## Chemistry - Gas Laws question

If propane is burnt in a limited amount of oxygen the poisonous carbon monoxide is formed as well as water vapour, so propane heaters must operate under conditions where oxygen supply is plentiful. When 5.35L of propane gas is burnt at a temperature of 273 K and pressure of 780mmHg in limited oxygen, what volume of carbon monoxide is produced at the same temperature and pressure?

I got an answer of 5.35L but apparently its wrong and i don't know. Can someone show how to work this out please

2. Have you written the equation for the reaction?

$C_3H_8 + \frac72O_2 \longrightarrow 3CO + 4H_2O$

So, if you burned 1 mol of propane you get 3 mol of carbon monoxide.

Or, if ou burn 1 litre of propane, you get 3 litres of carbon monoxide.

Can you solve your problem now?

3. Well, from the equation Unknown has produced, we use the gas law:

$PV=nRT$

where p is pressure, v is volume, r is 0.0821, t is temperature in kelvins, and n is the number of moles. Remember you must divide 780mmHg by 760, to get it into atm.

$1.026*5.35 = n(0.0821)(273)$

$n=0.245$ moles of propane.

Now, using the mole ratio, convert moles of propane to moles of carbon monoxide.

$(\frac{0.245mol}{1})(\frac{3molCO}{1mol})$

we get 0.735 moles of CO

now put this back into the gas law.

$1.026*V=(0.735)(0.0821)(273)$

then, after solving for v we get:

$V=16.06$liters of Carbon Monoxide.

4. Actually, you don't need to use the ideal gas equation since the conditions before and after the reaction are the same... you're only losing accuracy by using the ideal gas equation.

5. Originally Posted by Klutz
If propane is burnt in a limited amount of oxygen the poisonous carbon monoxide is formed as well as water vapour, so propane heaters must operate under conditions where oxygen supply is plentiful. When 5.35L of propane gas is burnt at a temperature of 273 K and pressure of 780mmHg in limited oxygen, what volume of carbon monoxide is produced at the same temperature and pressure?

I got an answer of 5.35L but apparently its wrong and i don't know. Can someone show how to work this out please
$\text{C}_3\text{H}_{8 \text{(l)}} + \frac{7}{2}\text{O}_{2\text{(g)}} \longrightarrow 3\text{CO}_{\text{(g)}} + 4\text{H}_2\text{O}_{\text{(g)}}$

The ideal gas equation is $PV = nRT$ and for our two conditions we have $P_1V_1 = n_1RT_1 \text{ and } P_2V_2 = n_2RT_2$

Since P and T are constant we can cancel them and dividing the second equation by the first one: $\dfrac{V_2}{V_1} = \dfrac{n_2}{n_1}$

Rearrange to give $V_2 = V_1 \cdot \dfrac{n_2}{n_1}$

$n_2$ and $n_1$ are given by the chemical reaction. $V_1$ is given in the question

edit: This post is essentially showing why Jerry's post (#2) holds true

edit 2: It seems unusual that the burning of propane (a strongly exothermic reaction) would take place under constant temperature