What would be the result if 5.0 L of hydrogen was burned with 2.0 l of oxygen in a closed container?
To determine the result of the reaction between hydrogen and oxygen in a closed container, we can use the balanced chemical equation for the combustion of hydrogen:
2H₂ + O₂ -> 2H₂O
This equation tells us that two molecules of hydrogen react with one molecule of oxygen to produce two molecules of water.
To calculate the result, we need to consider the stoichiometry of the reaction, which is the quantitative relationship between the reactants and products. The stoichiometry of this equation tells us that 2 moles of hydrogen react with 1 mole of oxygen to produce 2 moles of water.
To determine how many moles of hydrogen and oxygen are present, we need to convert the given volumes to moles using the Ideal Gas Law:
PV = nRT
Where:
P = pressure (assuming constant)
V = volume
n = number of moles
R = gas constant
T = temperature (assuming constant)
Since the question doesn't provide the pressure and temperature, we will assume that they are constant. Therefore, we can simplify the Ideal Gas Law to:
V = n
Now, let's convert the volumes to moles. We know that 1 mole of any gas occupies 22.4 liters at standard temperature and pressure (STP).
For hydrogen:
5.0 L * (1 mole / 22.4 L) = 0.223 moles of hydrogen
For oxygen:
2.0 L * (1 mole / 22.4 L) = 0.0893 moles of oxygen
Now that we know the number of moles for each reactant, we can determine the limiting reactant. The limiting reactant is the reactant that will be consumed first and, therefore, determines the amount of product formed.
To find the limiting reactant, we need to compare the moles of hydrogen and oxygen. The ratio of hydrogen to oxygen is 2:1, so we need to compare 0.223 moles of hydrogen to 0.0893 moles of oxygen.
Since we have more moles of hydrogen than oxygen, hydrogen is in excess, and oxygen is the limiting reactant.
Knowing this, we can calculate the maximum moles of water that can be formed using the stoichiometry of the balanced equation:
0.0893 moles of O₂ * (2 moles H₂O / 1 mole O₂) = 0.1786 moles of H₂O
Therefore, in a closed container, if 5.0 L of hydrogen was burned with 2.0 L of oxygen, the result would be the formation of 0.1786 moles of water.