A mixture containing 5.0 g of H2 and 10.0g of O 2 is allowed to react in a 15.0L container at a constant temperature of 25.0 C. Once the reaction is complete, what will be the pressure of the excess gas inside the container?

You must first determine which gas is in excess; i.e., which is the limiting reagent.

2H2 + O2 ==> 2H2O
mol H = g/molar mass = 5/2 = 2.5
mol O = 10/32 = 0.31

Using all of the H2 gas and all of the oxygen needed, how much H2O will be formed? That is
2.5 mol H2 x (2 mol H2O/2 mol H2) = 2.5 x 2/2 = 2.5 mol H2O form3ed.

Using all of the O2 gas and all of the hydrogen needed, how much H2O will be formed? That is
0.31 x (2 mol H2O/1 mol O2) = 0.31 mol H2O.

Obviously both of these answers can't be correct; therefore, in limiting reagent problems the SMALLER amount is the correct one and the reagent producing that value is the limiting reagent. Thus, O2 is the limiting reagent and hydrogen is in excess.

Next you want to determine how much H2 is used. That is
0.31 mol O2 x (2 mol H2/1 mol O2) = 0.31 x 2/1 = 0.62

We had 2.5 initially, we've used 0.62; therefore, we have 2.5-0.62 = about 1.9 remain unreacted. Use PV = nRT and solve for P (in atmospheres) using 298 for T in kelvin. I get an answer of about 3 atm. Check my work.

How many moles of hydrogen atoms are present in 2.41x10^24 molecules of CH4?

See your other question on CH3OH. The precess is the same.

DrBob222 -

You're the greatest!!!! Thank you!

To find the pressure of the excess gas inside the container after the reaction is complete, we need to determine which gas is the limiting reactant and calculate the amount of excess gas remaining.

Step 1: Convert the given masses of H2 and O2 to moles.
Molar mass of H2 = 2.02 g/mol
Molar mass of O2 = 32.00 g/mol

Moles of H2 = mass / molar mass = 5.0 g / 2.02 g/mol ≈ 2.48 mol
Moles of O2 = mass / molar mass = 10.0 g / 32.00 g/mol ≈ 0.31 mol

Step 2: Determine the stoichiometric ratio of the reactants.
From the balanced equation for the reaction between H2 and O2 to form H2O:
2H2 + O2 → 2H2O

The mole ratio is 2:1 between H2 and O2. This means that it takes 2 moles of H2 to react with 1 mole of O2.

Step 3: Identify the limiting reactant.
To determine which reactant is the limiting reactant, we compare the number of moles of each reactant to the stoichiometric ratio.

Since the stoichiometric ratio for H2 and O2 is 2:1, we have an excess of H2 because we have more than twice the amount (2.48 moles) needed to react with the available O2 (0.31 moles). Therefore, O2 is the limiting reactant.

Step 4: Calculate the moles of excess reactant left after the reaction.
The O2 is completely consumed in the reaction, leaving no excess O2.
The excess H2 can be calculated using the stoichiometric ratio:
2.48 moles H2 - (0.31 moles O2 × 2 moles H2 / 1 mole O2) = 2.16 moles H2

Step 5: Use the ideal gas law to calculate the pressure of the excess gas inside the container.
The ideal gas law is given by the equation:
PV = nRT

Where:
P = pressure in atm
V = volume in liters
n = number of moles of gas
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature in Kelvin

In this case, we are given:
V = 15.0 L
n = 2.16 mol
R = 0.0821 L·atm/(mol·K)
T = 25.0 + 273.15 K = 298.15 K (temperature converted to Kelvin)

Substituting these values into the ideal gas law equation, we can solve for P:
P × 15.0 L = 2.16 mol × 0.0821 L·atm/(mol·K) × 298.15 K

P = (2.16 mol × 0.0821 L·atm/(mol·K) × 298.15 K) / 15.0 L

Calculating this expression will give us the pressure of the excess gas inside the container.