1) Calculate the volume of air at 30 degrees C and 1.00 atm that is needed to burn completely 10.0 grams of propane. Assume that air is 21.0 percent O2 by volume.
2) Assuming all of the heat evolved in burning 30.0 g of propane is transferred to 8.00 kg of water (specific heat = 4.18 j/goK), calculate the increase in temperature of the water.
One problem per post please.
1. Write the balanced equation.
C3H8 + 5O2 ==> 3CO2 + 4H2O
Convert 10g C3H8 to mols, use the equation to convert mols C3H8 to mols O2, and use that and PV=nRT to calculate volume at the non-standard conditions. That divided by 0.21 should be the volume of air. Check my thinking. Check my work.
2. Different problem but same equation. Are you given the delta H for the reaction? If it isn't given then you must look it up in a set of tables. Use the value in the tables (usually given in kJ/mol) to calculate deltaH for the reaction. Then q=increase in T.
q = mass x specific heat x deltaT. Delta T will be the increase in T.
Post your work if you get stuck.
To clarify, q is not the increase in T; rather, q is the heat transferred to the water; q = increase in heat and that is used to calculate T.
Okay, for 2 I got the delta H, which is -103.7 kJ/mol. So is the equation something like:
(.682 mol C3H8)(-103.7kJ/mol)=(800g)(4.18J/g-K)(delta T)
and I solve for Delta T?
Oh, nevermind I figured this out, thanks DrBob222 for the help!
You're welcome! I'm glad you were able to figure it out. Just remember, if you ever need help with chemistry or any other topic, I'm here to assist you. And if you ever need a good laugh, well, I'm your bot! Just let me know what you need.
1) To calculate the volume of air needed to burn 10.0 grams of propane, we first need to convert the mass of propane to moles.
Given:
Mass of propane (C3H8) = 10.0 grams
Molar mass of propane (C3H8) = 44.11 g/mol
Number of moles of propane (C3H8) = Mass / Molar mass
= 10.0 g / 44.11 g/mol
= 0.2269 mol (rounded to 4 decimal places)
Using the balanced equation:
C3H8 + 5O2 → 3CO2 + 4H2O
We can see that 1 mole of propane (C3H8) reacts with 5 moles of oxygen (O2).
Number of moles of oxygen required = 5 x moles of propane
= 5 x 0.2269 mol
= 1.1345 mol (rounded to 4 decimal places)
Now, let's calculate the volume of air using the ideal gas law equation:
PV = nRT
Given:
Temperature (T) = 30 degrees Celsius + 273.15 K = 303.15 K (converted to Kelvin)
Pressure (P) = 1.00 atm
Number of moles of oxygen (n) = 1.1345 mol
Gas constant (R) = 0.0821 L·atm/mol·K
Rearranging the equation to solve for volume (V):
V = nRT / P
= (1.1345 mol)(0.0821 L·atm/(mol·K))(303.15 K) / 1.00 atm
= 28.08 L (rounded to 2 decimal places)
Since air is 21.0 percent O2 by volume, the required volume of air is:
Volume of air = Volume of O2 / 0.210
= 28.08 L / 0.210
= 133.71 L (rounded to 2 decimal places)
Therefore, the volume of air needed to burn completely 10.0 grams of propane at 30 degrees Celsius and 1.00 atm is approximately 133.71 liters.
Please let me know if I can assist you further.
For the first problem:
To calculate the volume of air needed to burn 10.0 grams of propane, you first need to write the balanced equation for the combustion reaction:
C3H8 + 5O2 --> 3CO2 + 4H2O
From the equation, you can see that 1 mole of propane (C3H8) requires 5 moles of oxygen (O2) to react completely.
1. Convert 10.0 grams of propane to moles:
10.0 g C3H8 * (1 mol C3H8 / molar mass of C3H8) = x mol C3H8
2. Use the stoichiometry of the balanced equation to determine the moles of oxygen required:
x mol C3H8 * (5 mol O2 / 1 mol C3H8) = y mol O2
3. Use the ideal gas law to calculate the volume of air (O2 + other gases) at the given temperature and pressure:
PV = nRT
P = 1.00 atm
T = 30°C + 273 = 303 K
R = 0.0821 L·atm/(mol·K)
(Volume of air) = (y mol O2 + other gases) * (0.0821 L·atm/(mol·K)) * (303 K) / (1.00 atm)
4. Finally, divide the volume of air by the fraction of O2 in air (21%) to determine the volume of air needed:
(Volume of air needed) = (Volume of air) / 0.21
Make sure to check if your thinking and work are correct.