A mixture containing 5.0 g of H2 and 10.0g of O 2 is allowed to react in a 15.0L container at a constant temperature of 25.0 C. Once the reaction is complete, what will be the pressure of the excess gas inside the container?
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To determine the pressure of the excess gas inside the container, we need to consider the reaction between hydrogen gas (H2) and oxygen gas (O2). The balanced chemical equation for this reaction is:
2H2 + O2 → 2H2O
From the balanced equation, we can see that the ratio of H2 to O2 needed for this reaction is 2:1. This means that each molecule of O2 needs two molecules of H2 to react completely.
To find out which reactant is in excess, we need to compare the number of moles of each substance present in the mixture.
First, let's calculate the moles of H2:
Molar mass of H2 = 2 g/mol
Number of moles of H2 = mass / molar mass = 5.0 g / 2 g/mol = 2.5 mol
Next, let's calculate the moles of O2:
Molar mass of O2 = 32 g/mol
Number of moles of O2 = mass / molar mass = 10.0 g / 32 g/mol = 0.3125 mol
Since the ratio of H2 to O2 in the balanced equation is 2:1, we can see that there is an excess of hydrogen gas.
Therefore, after the reaction is complete, all the O2 will be consumed, and only a portion of the H2 will be used. The remaining H2 will be the excess gas.
To calculate the pressure of the excess gas, we can use the ideal gas law:
PV = nRT
Where:
P is the pressure of the gas
V is the volume of the container (15.0 L)
n is the number of moles of the gas (the remaining H2)
R is the ideal gas constant (0.0821 L·atm/(mol·K))
T is the temperature in Kelvin (25.0°C + 273 = 298 K)
Let's substitute the values into the equation:
P * 15.0 L = 2.5 mol * 0.0821 L·atm/(mol·K) * 298 K
P * 15.0 L = 61.125 L·atm
Solving for P:
P = (61.125 L·atm) / 15.0 L ≈ 4.08 atm
Therefore, after the reaction is complete, the pressure of the excess gas (H2) inside the 15.0 L container will be approximately 4.08 atm.