1 gram of C3H8 gas and 1 gram of O2 gas are delivered to a metal sphere with a fixed volume of 1 L. After the two gases are introduced, the two reactants are ignited and burned according to the balanced reaction: C3H8(g)+5O2(g)changes to 3CO2(g)+4H2O(g)

After reaction, the sphere is held at 226.25 oC. What is the final pressure in the sphere given that all the compounds inside are gases and the limiting reagent is completely consumed with 100% yield ?

Now convert moles C3H8 to moles CO2. I have about 0.06.

Convert moles O2 to moles CO2. I have about 0.02.
Both answers can't be correct; the correct value in limiting reagent problems is ALWAYS the smaller value and the reagent providing that value is the limiting reagent. Therefore, O2 is the limiting reagent.
Since all of that will be consumed, you should convert moles O2 to moles C3H8 to determine moles C3H8 consumed, then subtract from initial moles to determine how much C3H8 remains unreacted. Add moles C3H8 + moles CO2 + moles H2O and use PV = nRT to solve for final P.

Laura, the second in line here is the response I made to the post last night. What don't you understand about it. Show your work if you are stuck.

i understood it when i was looking at the homework i did not check it so i reposed it. i did not mean to

Thanks for letting me know.

To find the final pressure in the metal sphere, we can apply the ideal gas law:

PV = nRT

Where:
P is the pressure
V is the volume (1 L in this case)
n is the number of moles of gas
R is the ideal gas constant (0.0821 L·atm/(mol·K))
T is the temperature in Kelvin (226.25°C = 499.4 K)

First, we need to determine the number of moles of each gas involved in the reaction.

Given:
1 gram of C3H8 gas (molar mass = 44 g/mol)
1 gram of O2 gas (molar mass = 32 g/mol)

To calculate the number of moles:
Number of moles = Mass of substance (in grams) / Molar mass

Moles of C3H8 = 1 g / 44 g/mol = 0.0227 moles
Moles of O2 = 1 g / 32 g/mol = 0.03125 moles

According to the balanced equation, one mole of C3H8 reacts with five moles of O2, producing three moles of CO2 and four moles of H2O.

Since C3H8 is the limiting reagent and is completely consumed, we can determine the moles of the products formed.

Moles of CO2 = 0.0227 moles C3H8 × (3 moles CO2 / 1 mole C3H8) = 0.0681 moles
Moles of H2O = 0.0227 moles C3H8 × (4 moles H2O / 1 mole C3H8) = 0.0908 moles

Now, let's calculate the total number of moles of gas present in the sphere after the reaction:

Total moles of gas = moles of CO2 + moles of H2O + moles of O2
= 0.0681 moles + 0.0908 moles + 0.03125 moles
= 0.19015 moles

Finally, we can substitute the values into the ideal gas law equation to find the final pressure:

P × V = n × R × T

P × 1 L = 0.19015 moles × (0.0821 L·atm/(mol·K)) × 499.4 K

P = (0.19015 × 0.0821 × 499.4) / 1
P ≈ 7.92 atm

Therefore, the final pressure in the metal sphere after the reaction is approximately 7.92 atm.