1. If 29.0 L of methane, CH4, undergoes complete
combustion at 0.961 atm and 140°C, how many liters
of each product would be present at the same
temperature and pressure?
2 If air is 20.9% oxygen by volume,
a. how many liters of air are needed for complete
combustion of 25.0 L of octane vapor, C8H18?
b. what volume of each product is produced?
3.Methanol, CH3OH, is made by causing carbon
monoxide and hydrogen gases to react at high
temperature and pressure. If 4.50 × 102 mL CO and
8.25 × 102 mL H2 are mixed,
a. which reactant is present in excess?
b. how much of that reactant remains after the
reaction?
c. what volume of CH3OH is produced, assuming the pressure and temperature are the same.
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answer
To answer these questions, we can use the concept of stoichiometry, which relates the balanced chemical equation to the quantities of reactants and products involved in a chemical reaction.
1. First, let's write the balanced chemical equation for the combustion of methane:
CH4 + 2O2 -> CO2 + 2H2O
According to the balanced equation, for each mole of methane, we need 2 moles of oxygen to completely react.
To find the number of moles of methane, we can use the ideal gas law equation:
PV = nRT
Here, we know the pressure (0.961 atm), volume (29.0 L), temperature (140°C), and the gas constant (R = 0.0821 L⋅atm/mol⋅K). We can rearrange the equation to solve for the number of moles (n) of methane:
n = PV / RT
Substituting the values, we get:
n = (0.961 atm) * (29.0 L) / (0.0821 L⋅atm/mol⋅K * 413.15 K)
Calculate n to find the number of moles of methane in 29.0 L.
Once we have the number of moles of methane, we can use the stoichiometry of the balanced equation to determine the number of moles of each product. Since the balanced equation tells us that 1 mole of methane produces 1 mole of carbon dioxide (CO2) and 2 moles of water (H2O), we can convert the moles of methane to moles of each product.
Finally, we can use the ideal gas law again to convert the moles of each product to volumes at the same temperature and pressure, using the formula:
V = nRT / P
Substitute the appropriate values to find the volume of each product.
2a. To determine the liters of air needed for complete combustion of octane vapor, C8H18, we can follow a similar process. First, you need to write the balanced chemical equation for the combustion of octane:
2C8H18 + 25O2 -> 16CO2 + 18H2O
According to the balanced equation, for each mole of octane, we need 25 moles of oxygen to completely react.
To find the number of moles of octane, you'll need to know the volume of octane vapor in liters and then use the ideal gas law as shown in question 1.
Once you have the number of moles of octane, you can use the stoichiometry to determine the number of moles of each product, similar to question 1.
2b. To determine the volume of each product produced, you can use the ideal gas law equation (V = nRT / P) once again. Convert the moles of each product to volumes at the same temperature and pressure.
3a. To determine which reactant is present in excess, we first need to find the number of moles of each reactant. Use the ideal gas law equation (PV = nRT) for both carbon monoxide (CO) and hydrogen (H2) gases. Convert the given volumes to moles using the balanced equation's cooefficients.
After finding the number of moles of each reactant, compare the two values to see which reactant is in excess. The reactant with a larger number of moles is present in excess.
3b. To find out how much of the excess reactant remains after the reaction, you'll need to use the stoichiometry of the balanced equation. Use the mole ratio between the excess reactant and the product (methanol, CH3OH) to determine the number of moles of the excess reactant consumed. Subtract this amount from the initial number of moles to find the remaining moles of the excess reactant.
3c. Finally, to determine the volume of methanol produced, you can use the ideal gas law equation (V = nRT / P) with the number of moles of methanol obtained from the stoichiometry calculation. Convert moles to volume using the given pressure and temperature.