How many liters of H2 are required to react completely with 5.7 liters of O2?
2H2 + O2 ==> 2H2O
When all gases are involved you can take a shortcut and NOT calculate mols. Use L directly.
5.7 L O2 x (2 mols H2/1 mol O2) = 5.7 x 2/1 = ?
To determine the number of liters of H2 required to react completely with 5.7 liters of O2, we need to consider the balanced chemical equation for the reaction between H2 and O2.
The balanced equation for this reaction is:
2H2 + O2 -> 2H2O
From this equation, we can see that 2 moles of H2 react with 1 mole of O2 to produce 2 moles of water.
To convert the given volume of O2 (5.7 liters) to moles, we need to use the ideal gas law equation:
PV = nRT
Where:
P = pressure of the gas (assume 1 atm)
V = volume of the gas (5.7 liters)
n = number of moles of the gas
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature of the gas (assume room temperature, around 298 K)
Rearranging the ideal gas law equation to solve for n:
n = PV / RT
n = 1 atm * 5.7 L / (0.0821 L·atm/mol·K * 298 K)
n = 0.22 moles of O2
Now, using the balanced equation, we know that 2 moles of H2 react with 1 mole of O2.
Therefore, the number of moles of H2 required to react completely with 0.22 moles of O2 is:
(2 moles H2 / 1 mole O2) * 0.22 moles O2 = 0.44 moles of H2
Finally, we need to convert the moles of H2 back to liters using the ideal gas law equation:
V = nRT / P
V = 0.44 moles * 0.0821 L·atm/mol·K * 298 K / 1 atm
V ≈ 11.9 liters
Therefore, approximately 11.9 liters of H2 are required to react completely with 5.7 liters of O2.