Water can be formed from its elements. what volume of oxygen would be required to react with 2.68L of Hydrogen at STP?
2H2 + O2 => 2H2O
2.68L H2 x (1 mol O2/2 mols H2) = ?L O2.
To find out the volume of oxygen needed to react with a certain volume of hydrogen at STP (Standard Temperature and Pressure), we need to use the balanced chemical equation for the reaction between hydrogen and oxygen to form water.
The balanced chemical equation is as follows:
2H₂ + O₂ → 2H₂O
From the equation, we can see that it takes two moles of hydrogen for every one mole of oxygen to produce two moles of water.
First, we need to determine the number of moles of hydrogen in 2.68 L of hydrogen. To do this, we will use the ideal gas law equation, which is:
PV = nRT
Where:
P = pressure (STP is 1 atm)
V = volume (2.68 L)
n = number of moles (what we need to calculate)
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature (273.15 K at STP)
Rearranging the equation to solve for n, we have:
n = PV / RT
Substituting the values, we get:
n = (1 atm) * (2.68 L) / ((0.0821 L·atm/(mol·K)) * (273.15 K))
n ≈ 0.1108 moles of hydrogen
Since the balanced equation shows that two moles of hydrogen react with one mole of oxygen, we need half the moles of oxygen compared to hydrogen.
Thus, the moles of oxygen required would be:
0.1108 moles of hydrogen / 2 = 0.0554 moles of oxygen
Now, to find the volume of oxygen needed, we can use the ideal gas law equation again, using the moles of oxygen calculated:
PV = nRT
Rearranging the equation to solve for V, we have:
V = nRT / P
Substituting the values, we get:
V = (0.0554 moles) * (0.0821 L·atm/(mol·K)) * (273.15 K) / (1 atm)
V ≈ 1.44 L
Therefore, approximately 1.44 liters of oxygen would be required to react with 2.68 liters of hydrogen at STP to form water.