Assuming the volumes of all gases in the reaction are measured at the same temperature and pressure, calculate the volume of water vapor obtainable by the explosive reaction of a mixture of 725mL of hydrogen gas and 325mL of oxygen gas.
andd...
Calculate the volume of air at 30 degrees C and 1.00atm that is needed to burn completely 10.0 grams of propane. Assume that air is 21.0% oxygen (O2) by volume.
2H2 + O2 ==> 2H2O
This may be solved the usual stoichiometric way by changing g H2 and g O2 to mols, determining which is the limiting reagent, etc etc. OR, the shortcut way is to simply use volumes since all are gases. If you don't get this way, I can explain how it works in detail. Let me know.
To use volumes only, 725 mL H2 will form 725 mL H2O since the ratio of H2 to H2O is 2:2 (or 1:1). 325 mL O2 will form 325 x (2 mols H2O/1 mol O2) = 650 mL H2O. Therefore, 650 mL H2O will be the final product (O2 is the limiting reagent). There will be excess H2 remaining unreacted.
For #2, write the equation and balance it.
C3H8 + 5O2 ==> 3CO2 + 4H2O
Convert 10 g propane to mols. #mols = grams/molar mass propane.
Convert mols Propane to mols O2 using the coefficients in the balanced equation.
Use PV = nRT to convert mols O2 to volume O2 at the conditions cited.
Convert to volume of air by the following:
volume oxygen at cited conditions/0.21 = volume air.
Post your work if you get stuck or if you have any questions.
another question...
I have a problem that states:
Find the pressure of a sample of carbon tetrachloride, CCl4 if 1.00 mol occupies 35.0L at 77.0 degress C (slightly above its normal boiling point). Assume that CCl4 obeys (a) the ideal gas law; (b) the van der Waals equation. The van der Waals constants for CCl4 are a= 20.39(L^2)(atm)/mol^2 and b= 0.1383L/mol.
I tried to do the problems and the answers I got are (a)=.814atm and (b)=.800atm but the back of the book is telling me that (a)=0.821atm an (b)=0.805. I'm not sure if i'm wrong or if the book is off.
To calculate the volume of water vapor obtainable by the explosive reaction of hydrogen gas and oxygen gas, we need to determine the mole ratio of hydrogen to oxygen in the reaction and then use it to calculate the volume of water vapor produced.
1. Determine the number of moles of hydrogen gas and oxygen gas:
- Convert the volume of hydrogen gas from mL to L:
725 mL ÷ 1000 mL/L = 0.725 L
- Convert the volume of oxygen gas from mL to L:
325 mL ÷ 1000 mL/L = 0.325 L
2. Convert the volumes of hydrogen gas and oxygen gas to moles using the ideal gas law:
- Moles of hydrogen gas (H2):
Moles of H2 = Volume of H2 / molar volume of H2
= 0.725 L / 22.4 L/mol
- Moles of oxygen gas (O2):
Moles of O2 = Volume of O2 / molar volume of O2
= 0.325 L / 22.4 L/mol
3. Determine the mole ratio of hydrogen to oxygen in the reaction. Assuming the balanced chemical equation for the reaction is 2H2 + O2 → 2H2O, the mole ratio is 2:1.
4. Calculate the moles of water vapor produced:
- Moles of water vapor = Moles of oxygen gas / 2
5. Calculate the volume of water vapor produced using the ideal gas law:
- Volume of water vapor = Moles of water vapor *molar volume of water vapor
Now, let's move to the second question.
To calculate the volume of air needed to burn propane, we'll follow these steps:
1. Convert the mass of propane to moles:
- Propane (C3H8) has a molar mass of 44.1 g/mol.
- Moles of propane = Mass of propane / Molar mass of propane
2. Calculate the moles of oxygen required to burn the propane:
- The balanced equation for the combustion of propane is C3H8 + 5O2 -> 3CO2 + 4H2O.
- From the equation, we can see that 1 mole of propane requires 5 moles of oxygen.
- Moles of oxygen needed = Moles of propane * 5
3. Calculate the volume of air needed, assuming air is 21.0% oxygen by volume:
- The mole ratio of oxygen to air is 0.21 (21.0% oxygen).
- Volume of air = Moles of oxygen / Mole ratio of oxygen to air
4. Convert the volume of air to liters using the ideal gas law if necessary.
Note: In both calculations, make sure to use the same temperature and pressure units throughout the calculations.
To calculate the volume of water vapor obtainable by the explosive reaction of hydrogen and oxygen gases, we can use the balanced chemical equation for the reaction:
2H₂(g) + O₂(g) → 2H₂O(g)
1. Start by converting the given volumes of hydrogen and oxygen gases to moles. To do this, divide the volume by the molar volume at the given conditions (same temperature and pressure). The molar volume at standard temperature and pressure (STP) is approximately 22.4 L/mol.
For hydrogen gas:
Volume = 725 mL = 725 / 1000 L = 0.725 L
Moles of H₂ = Volume of H₂ / Molar volume at given temperature and pressure
For oxygen gas:
Volume = 325 mL = 325 / 1000 L = 0.325 L
Moles of O₂ = Volume of O₂ / Molar volume at given temperature and pressure
2. Once you have the moles of hydrogen and oxygen gases, use the stoichiometry of the balanced equation to determine the moles of water vapor that can be produced. From the balanced equation, we can see that 2 moles of hydrogen gas react with 1 mole of oxygen gas to produce 2 moles of water vapor.
3. Finally, we can convert the moles of water vapor to volume using the molar volume at the given conditions.
For the equation 2H₂(g) + O₂(g) → 2H₂O(g), we can say that the volume of water vapor is equal to the moles of water vapor times the molar volume at the given temperature and pressure.
To calculate the volume of air needed to burn propane, we will use the balanced chemical equation for the combustion of propane:
C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(g)
1. Start by converting the given mass of propane to moles. To do this, divide the mass by the molar mass of propane (C₃H₈), which is approximately 44.1 g/mol.
2. Next, use the balanced equation to determine the mole ratio between propane and oxygen. From the balanced equation, we can see that 1 mole of propane reacts with 5 moles of oxygen.
3. Once you have the moles of propane, use the mole ratio to determine the moles of oxygen required. Multiply the mole ratio by the moles of propane.
4. Then, calculate the moles of air needed by dividing the moles of oxygen by the fraction of oxygen in air (21.0%). This accounts for the fact that air is not pure oxygen. Multiply the moles of oxygen by the reciprocal of the fraction of oxygen in air (i.e., divide by 0.21).
5. Finally, convert the moles of air to volume using the molar volume at the given temperature and pressure. The molar volume at STP (standard temperature and pressure) is approximately 22.4 L/mol.
For the equation C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(g), we can say that the volume of air is equal to the moles of air times the molar volume at the given temperature and pressure.