How many liters of oxygen are required to react completely with 2.4 liters of hydrogen to form water? H2 + O2--> H2O?
Balance the equation:
2H2+O2 = 2H2O
So, you have half as many moles of O2 as of H2
So, 1/2 the volume.
To determine the number of liters of oxygen required to react completely with 2.4 liters of hydrogen to form water, we need to start by balancing the chemical equation.
The balanced chemical equation for the reaction between hydrogen gas (H2) and oxygen gas (O2) to form water (H2O) is:
2H2 + O2 --> 2H2O
From the balanced equation, we can see that 2 moles of hydrogen gas react with 1 mole of oxygen gas to form 2 moles of water.
Since we are given the volume in liters, we need to convert it to moles using the ideal gas law. The ideal gas law is expressed as:
PV = nRT
Where:
P = pressure (constant at constant temperature and volume)
V = volume
n = number of moles
R = ideal gas constant
T = temperature
Since we are working with gases at room temperature and assuming constant pressure, we can simplify the equation to:
V = nRT/P
Now, let's calculate the number of moles of hydrogen gas using the given volume:
V = 2.4 L (volume of hydrogen gas)
To convert this volume to moles, we need to determine the value of the gas constant, R, and the temperature, T. At standard temperature and pressure (STP), R is approximately 0.0821 L·atm/(K·mol) and T is 273.15 K.
Plugging in the values, we have:
n(H2) = V/RT
= 2.4 L / (0.0821 L·atm/(K·mol) * 273.15 K)
≈ 0.105 moles of H2
From the balanced chemical equation, we know that 2 moles of hydrogen react with 1 mole of oxygen. Therefore, we can calculate the number of moles of oxygen required:
n(O2) = (n(H2)) / 2
≈ 0.105 moles / 2
≈ 0.0525 moles of O2
Now, let's convert this back into volume:
V(O2) = n(O2) * RT/P
≈ 0.0525 moles * (0.0821 L·atm/(K·mol) * 273.15 K) / 1 atm
≈ 1.123 L of O2
Therefore, approximately 1.123 liters of oxygen are required to react completely with 2.4 liters of hydrogen to form water.