Please help and explain each problem.
PLEASE!!!!
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
same pressure?
I'll be glad to help if you want to help yourself but I'm not going to do all of your work. What don't you understand? You say you have tried but can't get the right answer. So show what you've done so we can get an idea where you're going wrong. You could have done it by now with the time you've wasted reposting and waiting.
Sure, I'd be happy to help and explain each problem!
1. In order to solve this problem, we can use the ideal gas law to find the number of moles of methane and then use stoichiometry to determine the volumes of the products. Here are the steps to solve this problem:
Step 1: Convert the given temperature from Celsius to Kelvin by adding 273.15. So, 140°C = (140 + 273.15) K = 413.15 K.
Step 2: Use the ideal gas law equation PV = nRT to find the number of moles of methane (CH4). Rearrange the equation to solve for n (number of moles):
n = PV / RT
Where P is the pressure in atm, V is the volume in liters, R is the ideal gas constant (0.0821 L.atm/mol.K), and T is the temperature in Kelvin.
Step 3: Plug in the given values into the equation. Given: P = 0.961 atm and V = 29.0 L. R is already provided.
n = (0.961 atm * 29.0 L) / (0.0821 L.atm/mol.K * 413.15 K)
Calculate n to get the number of moles of methane.
Step 4: Use the balanced chemical equation for complete combustion of methane to determine the stoichiometric ratios. The balanced equation is:
CH4 + 2O2 → CO2 + 2H2O
This means that for every 1 mole of methane, we get 1 mole of carbon dioxide (CO2) and 2 moles of water (H2O).
Step 5: Calculate the number of moles of carbon dioxide (CO2) and water (H2O) produced using the stoichiometric ratios obtained from the balanced chemical equation. Multiply the number of moles of methane by the stoichiometric coefficients of carbon dioxide and water.
Step 6: Finally, use the ideal gas law equation to find the volumes of carbon dioxide and water at the same temperature and pressure. Rearrange the equation to solve for V (volume):
V = nRT / P
Where V is the volume in liters, n is the number of moles, R is the ideal gas constant, T is the temperature in Kelvin, and P is the pressure in atm.
Plug in the values of n, R, T, and P to calculate the volumes of carbon dioxide and water.
2. For this problem, we need to determine the volumes of air required for complete combustion of octane vapor (C8H18) and the volumes of each product produced. To solve these questions, we can use the stoichiometry of the combustion reaction of octane.
a. To find the volume of air required, we first need to calculate the number of moles of octane (C8H18). Use the ideal gas law equation PV = nRT to find the number of moles of octane, just like in problem 1. Rearrange the equation to solve for n:
n = PV / RT
Plug in the given values of P, V, R, and T to calculate the number of moles of octane.
b. Once we know the number of moles of octane, we can use the balanced chemical equation for the combustion reaction to determine the stoichiometric ratios. The balanced equation is:
C8H18 + 12.5O2 → 8CO2 + 9H2O
According to the balanced equation, for every 1 mole of octane, we need 12.5 moles of oxygen (from air), and we get 8 moles of carbon dioxide and 9 moles of water.
To find the volume of each product, we can use the ideal gas law equation V = nRT / P. Plug in the calculated number of moles for each product, the given temperature and pressure, and the ideal gas constant to calculate the volumes.
3. To solve this problem, we need to determine which reactant is present in excess, how much of that reactant remains after the reaction, and the volume of methanol (CH3OH) produced.
a. To determine which reactant is in excess, we need to compare the moles of carbon monoxide (CO) and hydrogen gas (H2) given. Use the ideal gas law equation PV = nRT to find the number of moles for both reactants. Rearrange the equation to solve for n:
n = PV / RT
Plug in the given values of pressure, volume, temperature, and the ideal gas constant to calculate the number of moles for each reactant.
b. Once we have the number of moles for each reactant, compare their mole ratios from the balanced chemical equation. The balanced equation is:
CO + 2H2 → CH3OH
According to the balanced equation, 1 mole of carbon monoxide reacts with 2 moles of hydrogen gas to produce 1 mole of methanol. Compare the moles of CO and H2 to determine which one is present in excess.
c. To find the volume of methanol produced, we can use the ideal gas law equation V = nRT / P. Rearrange the equation to solve for V:
V = nRT / P
Plug in the number of moles of methanol (which is equal to the number of moles of CO used in the reaction), the given temperature and pressure, and the ideal gas constant to calculate the volume.