Verified answer

Hello there! :)

Here's how I did it:

We have the equation:

C₃H₈ (g) + 5O₂ (g) → 3CO₂ (g) + 4H₂O (g)

So far, we know:

525 g O₂

523 K (CO₂) <-- (K = °C + 273. => K = 250°C + 273 = 523 K)

7.25 atm (CO₂)

The first step is to convert the given amount of oxygen, which is in grams, into moles of oxygen.

To do this, we use the molar mass of oxygen and dimensional analysis:

525 g O₂ * (1 mol O₂ / 32.0 g O₂) = 16.41 mol O₂

Thus, there is 16.41 moles of oxygen gas in 525 grams of oxygen gas.

Now, we calculate the number of moles of CO₂ gas that forms, given that the mole ratio between oxygen and carbon dioxide is 5:3 and that we have 16.41 moles of oxygen gas:

16.41 mol O₂ * (3 mol CO₂ / 5 mol O₂) = 9.85 mol CO₂

Thus, 9.85 moles of CO₂ will form from 16.41 moles of O₂

Now, we use the ideal gas law equation to calculate the volume:

PV = nRT

We want to solve for volume. Therefore, divide both sides by P to isolate V:

V = (nRT) / P

Now, we plug in our given information and the information that we found out to calculate V:

n = 9.85 mol CO₂

R = 0.08205746 (L * atm) / (K * mol)

T = 523 K

P = 7.25 atm

V = (nRT) / P

V = (9.85 * 0.08205746 * 523) / (7.25)

V = 58.31 L CO₂

Therefore, 58.31 litres of CO₂ will form from 525 grams of oxygen gas.

Hope this helps! :)

Step 1. Convert the masses into moles

moles of O2 = 525 / 32 = 16.4 moles of O2

Assuming unlimited C3H8 then you would expect to produce 16.04 x 2/5 moles of CO2 which equals 6.56 moles of CO2.

Now use V = nRT / P to find the volume, where n = 6.56 and T = (250 + 273). The rest is up to you.