How many grams of propane (C3 H8 ) will react with 3.29 L of O at 1.05 atm and -34°C?
i know the balanced eqn is 1C3H8 + 5 O2 -----> 3CO2 + 4H20
and that PV=nrt , with T= -34 + 273.15 = 239.15K.
n=(PV)/(RT) (F=for O2). = 0.176 mol O2
0.176 mol O2 * (1mol C3H8 / 5mol O2) * 44.0962gC3H8/1mol C3H8)
I get 1.552 g of C3H8 react with O2 but i feel that its incorrect and was wondering if i iwas doing it properly
Not only have you solved the problem correctly but you have arrived at the correct answer.
You are on the right track with your calculations. Let's break it down step by step:
1. Find the moles of oxygen (O2):
Using the ideal gas law equation, PV = nRT, we can rearrange it to solve for moles (n):
n = (PV) / (RT)
Given that P = 1.05 atm, V = 3.29 L, R = 0.0821 L·atm/(K·mol), and T = -34°C + 273.15 = 239.15 K, we can substitute the values:
n = (1.05 atm * 3.29 L) / (0.0821 L·atm/(K·mol) * 239.15 K)
n = 0.1762 mol
2. Use the stoichiometry of the balanced equation to find the moles of propane (C3H8):
According to the balanced equation, 1 mole of propane reacts with 5 moles of oxygen. Therefore, we need to convert the moles of oxygen to moles of propane:
0.1762 mol O2 * (1 mol C3H8 / 5 mol O2) = 0.03524 mol C3H8
3. Finally, convert moles of propane to grams of propane:
To convert moles of propane (C3H8) to grams, we need to multiply the moles by the molar mass of propane. The molar mass of C3H8 is 44.0962 g/mol:
0.03524 mol C3H8 * 44.0962 g/mol = 1.552 g of C3H8
So your calculation is correct. The reaction will require approximately 1.552 grams of propane (C3H8) to react with 3.29 L of oxygen (O2) at 1.05 atm and -34°C.
To determine the number of grams of propane (C3H8) that will react with a given amount of O2, you have correctly identified the balanced equation and calculated the number of moles of O2.
First, convert the volume of O2 to moles using the ideal gas law equation:
n(O2) = (PV) / (RT)
n(O2) = (1.05 atm * 3.29 L) / (0.0821 L·atm/(mol·K) * 239.15 K)
n(O2) = 0.1772 mol O2
Next, use the stoichiometry of the balanced equation to determine the moles of propane that react with the moles of O2:
n(C3H8) = n(O2) * (1 mol C3H8 / 5 mol O2)
n(C3H8) = 0.1772 mol O2 * (1 mol C3H8 / 5 mol O2)
n(C3H8) = 0.03544 mol C3H8
Finally, calculate the grams of propane using the molar mass of C3H8:
mass(C3H8) = n(C3H8) * molar mass(C3H8)
mass(C3H8) = 0.03544 mol C3H8 * 44.0962 g/mol C3H8
mass(C3H8) = 1.5614 g C3H8
Therefore, the correct answer is approximately 1.5614 grams of propane (C3H8) will react with 3.29 L of O2 at 1.05 atm and -34°C.