How many litres of water are produced when 3.3 litres of hydrogen is burnt in excess of oxygen at 25degree centigrade?
To determine the number of liters of water produced when 3.3 liters of hydrogen is burned in excess oxygen, we need to understand the chemical equation for the reaction.
The balanced chemical equation for the reaction of hydrogen and oxygen to produce water is:
2H₂ + O₂ → 2H₂O
From the equation, we can see that 2 moles of hydrogen react with 1 mole of oxygen to produce 2 moles of water.
To calculate the number of moles of hydrogen in 3.3 liters, we need to use the ideal gas law equation:
PV = nRT
Where:
P = pressure (assuming it is at standard pressure, which is around 1 atm)
V = volume of the gas (3.3 L)
n = number of moles
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature in Kelvin (25°C + 273.15 = 298.15 K)
Rearranging the ideal gas law equation to solve for n, we have:
n = PV / RT
Substituting the values, we get:
n = (1 atm) * (3.3 L) / (0.0821 L·atm/(mol·K)) * (298.15 K)
Simplifying the equation:
n = 0.133 moles of hydrogen
From the balanced equation, we know that 2 moles of hydrogen react with 2 moles of water. Therefore, 0.133 moles of hydrogen will produce 0.133 moles of water.
To convert moles to liters, we need to use the ideal gas law equation again:
V = nRT / P
Substituting the values:
V = (0.133 moles) * (0.0821 L·atm/(mol·K)) * (298.15 K) / (1 atm)
Simplifying the equation:
V ≈ 3.37 liters of water
Therefore, when 3.3 liters of hydrogen is burned in excess oxygen at 25 degrees Celsius, approximately 3.37 liters of water are produced.