What would be the result if 5.0 L of hydrogen were burned with 2.0 L of oxygen in a closed containner?
2H2 + O2 ==> 2H2O
First determine the limiting reagent. (When BOTH reactants are given you know it is a limiting reagent problem.)
5 L H2 x (2 moles H2O/2 moles H2) = 5 L x (2/2) = 5 L H2O.
2 L O2 x (2 moles H2O/1 mole O2) = 4 L H2O.
Both answers can be right; the correct answer in limiting reagent problems is ALWAYS the smaller value.
Some of the H2 will remain un-reacted.
To determine the result of burning hydrogen with oxygen in a closed container, we need to know the balanced chemical equation for the reaction between hydrogen and oxygen. The balanced equation is:
2H₂ + O₂ → 2H₂O
This equation tells us that two molecules of hydrogen react with one molecule of oxygen to produce two molecules of water.
Next, we need to calculate the limiting reactant. The limiting reactant is the one that is completely consumed first and determines the amount of product formed. To find the limiting reactant, we compare the number of moles of each reactant.
First, we convert the volumes of hydrogen and oxygen to moles using the ideal gas law:
Number of moles = (volume of gas) / (molar volume at STP)
The molar volume at STP (Standard Temperature and Pressure) is 22.4 L/mol.
Number of moles of hydrogen = 5.0 L / 22.4 L/mol = 0.2232 mol
Number of moles of oxygen = 2.0 L / 22.4 L/mol = 0.0893 mol
According to the balanced equation, we need twice as many moles of hydrogen as oxygen to react completely. This means that the oxygen is the limiting reactant because we have fewer moles of oxygen compared to hydrogen.
Therefore, all 0.0893 mol of oxygen will react with 0.0446 mol (0.0893 mol / 2) of hydrogen.
Using the stoichiometry of the balanced equation, we can determine the amount of water produced. Since 2 moles of water are produced for every 2 moles of hydrogen reacted, the amount of water produced will also be 0.0446 mol.
Finally, we convert the moles of water to volume using the molar volume at STP:
Volume of water produced = (moles of water) x (molar volume at STP)
= 0.0446 mol x 22.4 L/mol
≈ 0.998 L
Therefore, if 5.0 L of hydrogen and 2.0 L of oxygen were burned completely in a closed container, approximately 0.998 L of water would be produced.