What volume of hydrogen is necessary to react with five liters of nitrogen to produce ammonia?
N2 + 3H2 --> 2NH3
The equation is balanced.
1 liter of N2/ 3 liters of H2
5 liters of N2/ x liters of H2
Set the two fraction equal and cross multiply to solve for x.
To determine the volume of hydrogen required to react with five liters of nitrogen, we need to use the stoichiometry of the reaction.
From the balanced equation:
1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3.
Given that the volume of nitrogen is five liters, we need to convert it to moles using the ideal gas law:
PV = nRT
As the pressure, temperature, and gas constant are not provided, we can assume standard conditions, which are 1 atm and 273.15 K.
Using the ideal gas law equation:
PV = nRT
1 atm * 5 L = n * 0.0821 atm L/mol K * 273.15 K
5 atm L = 22.414 L/mol * n
Now, we can find the number of moles of nitrogen:
n(N2) = 5 atm L / 22.414 L/mol = 0.2236 mol
According to the stoichiometry of the reaction, 1 mole of nitrogen requires 3 moles of hydrogen. Therefore, to find the volume of hydrogen required, we multiply the number of moles of nitrogen by the stoichiometric ratio.
n(H2) = 0.2236 mol N2 * (3 mol H2 / 1 mol N2) = 0.6708 mol H2
Now, we can convert moles of hydrogen to liters by using the ideal gas law:
PV = nRT
1 atm * V = 0.6708 mol * 0.0821 atm L/mol K * 273.15 K
V = (0.551 L)
Therefore, the volume of hydrogen required to react with five liters of nitrogen to produce ammonia is approximately 0.551 liters.