5.00 g H2 and 10.0 g O2 are combined in a 10.0 L container at 25°C and then ignited. Calculate the final pressure in the container. You must compare the pressure the water would have if it all existed as vapor to the vapor pressure of water at that temperature! If greater it condenses to have the equilibrium vapor pressure.

Question

5.00 g H2 and 10.0 g O2 are combined in a 10.0 L container at 25°C and then ignited. Calculate the final pressure in the container. You must compare the pressure the water would have if it all existed as vapor to the vapor pressure of water at that temperature! If greater it condenses to have the equilibrium vapor pressure.

Answer 1

Sorry i posted the answer to a different question:

What volume of carbon dioxide would be produced at 350˚C and
1.00 atm from the complete thermal decomposition of 1.00 kg of
magnesium carbonate, MgCO3? What is the density of carbon dioxide
at this temperature?

Answer 2

To calculate the final pressure in the container after the reaction, we need to consider the chemical reaction involved and the gas laws.

First, let's write the balanced chemical equation for the reaction between hydrogen (H2) and oxygen (O2) to form water (H2O):

2H2 + O2 -> 2H2O

According to the stoichiometry of the reaction, every 2 moles of H2 and 1 mole of O2 will produce 2 moles of H2O. So, let's calculate the number of moles of H2 and O2 given the masses provided:

Molar mass of H2 = 2 g/mol
Molar mass of O2 = 32 g/mol

Number of moles of H2 = mass of H2 / molar mass of H2
= 5.00 g / 2 g/mol
= 2.50 mol

Number of moles of O2 = mass of O2 / molar mass of O2
= 10.0 g / 32 g/mol
= 0.313 mol

Since the reaction is balanced on a mole basis, the ratio of moles for H2 and O2 is 2:1. Therefore, O2 is the limiting reactant, and all of the O2 will be consumed in the reaction.

Now, let's consider the ideal gas law, which states:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.

Given that the volume of the container is 10.0 L and the temperature is 25°C, let's convert the temperature to Kelvin:

T(K) = T(°C) + 273.15
= 25°C + 273.15
= 298.15 K

Now, we can calculate the pressure due to the water vapor using the vapor pressure of water at 25°C. The vapor pressure of water at 25°C is approximately 23.8 mmHg.

If the pressure due to the water vapor is greater than 23.8 mmHg, the excess water vapor will condense to reach equilibrium with the vapor pressure. Therefore, the final pressure in the container will be equal to the vapor pressure of water.

If the pressure due to the water vapor is less than or equal to 23.8 mmHg, it means there is no excess water vapor, and the final pressure will be the sum of the pressure due to the water vapor and the remaining gases in the container.

To calculate the pressure due to the water vapor, we can use Dalton's Law of Partial Pressures. According to this law, the total pressure (PTotal) is equal to the sum of the partial pressures of each gas:

PTotal = PWater Vapor + PRemaining Gases

Since all the oxygen gas reacts and is consumed, the pressure due to the remaining gases is only dependent on the pressure due to the water vapor. So:

PTotal = 2 * PWater Vapor

Now, let's plug in the values:

PTotal = 2 * 23.8 mmHg
= 47.6 mmHg

Therefore, the final pressure in the container is 47.6 mmHg.

Answer 3

MgCO3 --> MgO + CO2

1.00kg MgCO3 x 10^3g/1kg x 1 mol MgCO3/84.32g MgCO3 x 1 mol CO2/1 mol MgCO3 = 11.860 mol CO2

PV=nRT
V = nRT/P = 11.860 mol x 0.08206 Latm/molK x 632K / 1.00 atm = 606.32L

1.00kg MgCO3 x 10^3g/1kg x 1 mol MgCO3/84.32g MgCO3 x 1 mol CO2/1 mol MgCO3 x 44.01g CO2/1 mol CO2 = 522g CO2

522g/606.32L = 861 g/mL