GAS STOICHIMETRY

Question

1. If 29.0L of octane, C9 H20, undergoes complete combustion at 0.961 atm and 200 C, how many liters of each gaseous product (carbon dioxide gas and water vapor gas) are formed?
2. What mass of potassium chloride can be collected at 0.987 atm of pressure and 250 C when 230.6g of KClO3 goes through the following reaction?

KClO3 (s) KCl (s) + O2 (g)

Answer 1

Hi there Hetvi!

In these two problems, we're covering and merging concepts of reactions with the Ideal Gas Law, PV = nRT.

Question 1:

Stoichiometry → Ideal Gas Law for octane → Mole Ratios → Ideal Gas Law for Products

Complete combustion reactions are always defined as:

Fuel + O2 → CO2 + H2O (+ Heat) ; for some ratios of fuel and O2 to CO2 and water. Fuel reacts with O2 to form CO and H2O, then CO can react further to become CO2. Meanwhile, liquid water quickly boils to become steam.

We know our fuel is octane, C9H20, so we must balance the reaction using stoichiometry:

nC9 H20 + xO2 → y CO2 + z H2O

Start with n = 1 and y = 9 to balance carbons first since there's only one place they can go.

C9 H20 + xO2 → 9 CO2 + z H2O

Then let's work make z = 10 for hydrogen for the same reason.

C9 H20 + xO2 → 9 CO2 + 10 H2O

Now, we know we end with 18+10 oxygens so x = 14.

C9 H20 + 14 O2 → 9 CO2 + 10 H2O

Now we know our mole ratios for the combustion reaction. For every 1 molecule of octane combusted with 14 molecules of O2, we produce 9 molecules of CO2 and 10 molecules of water. Of course, this can scale up to 10 or 20,000 molecules or entire moles - huge amounts of molecules. The ratios stay the same.

Looking up octane's boiling point, we find that octane is a vapor above 125.6 °C at sea level (P = 1 atm). We're at even less pressure so the boiling point is even lower. (I have a strong Thermodynamics background.) So, we can apply the Ideal Gas Law to find what amount of octane we have:

P = 0.961 atm

T = 200 °C = 473.15 K (add 273.15 to convert °C to K)

R = 0.08206 L • atm / (mol • K)

V = 29.0 L

Before plugging in, let's rearrange to solve for n:

PV = nRT

PV / (RT) = n

0.961 atm * 29.0 L / (0.08206 L • atm / (mol • K) * 473.15 K)

= 0.718 mol octane

So now we can use our molar ratios to find moles of the products:

n_CO2 = 9 * n_octane = 6.46 mol CO2

n_H2O = 10 * n_octane = 7.18 mol H2O

Finally, we apply the Ideal Gas Law again to bring moles back to liters:

PV = nRT

V = nRT / P

Assuming pressure and temperature are forced to remain the same (they would both increase, so we would have to expand the container and cool the gas)

P = 0.961 atm

T = 473.15 K

V_CO2 = 6.46 mol * 0.08206 L • atm / (mol • K) * 473.15 K / 0.961 atm

= 261 L CO2

V_H2O = 7.18 mol * 0.08206 L • atm / (mol • K) * 473.15 K / 0.961 atm

= 290 L H2O

(A quick trick would be to use n1V1 = n2V2 for each, since T and P remain constant.)

Question 2:

Mass → moles → moles → mass

via Molar Mass → Stoichiometry → Molar Mass

A quick balance shows that:

KClO3 (s) → KCl (s) + 1½ O2 (g)

So KCl is formed at a 1:1 ratio to KClO3. We can find that the molar mass of KClO3 and KCl are:

MM_K = 39.1 g/mol

MM_Cl = 35.45 g/mol

MM_O = 16.0 g/mol

MM_KCl = MM_K + MM_Cl = 39.1 g/mol + 35.45 g/mol = 74.55 g/mol

MM_KClO3 = MM_KCl + 3 * MM_O = 74.55 g/mol + 3 * 16.0 g/mol = 122.6 g/mol

Now, assuming a complete reaction (no leftover KClO3), we can convert mass of KClO3 to moles, then to moles of KCl, and finally to mass of KCl.

m_KClO3 = 230.6 g

m = n * MM

m / MM = n

n_KClO3 = 230.6 g / 122.6 g/mol = 1.88 mol KClO3

n_KClO3 = n_KCl

m_KCl = 1.88 mol * 74.55 = 140.3 g KCl.

Hope this helps! Please let me know if you have any questions!