What volume of ammonia at 7.0 atm and 235C would you get from 333g of nitrogen reactin with enough mass of hydrogen?k
calculate the density of hydrogen at STP?
Write the equation and balance it. Then use this link to set up the calculation of moles N2; finally use PV = nRT to convert moles N2 at STP (from the calculation) to the conditions in the problem.
http://www.jiskha.com/science/chemistry/stoichiometry.html
density H2 at STP = molar mass H2/22.4
To find the volume of ammonia produced, we first need to calculate the number of moles of nitrogen, the number of moles of hydrogen, and then determine the limiting reagent.
1. Calculate the number of moles of nitrogen:
The molar mass of nitrogen (N₂) is 28.02 g/mol.
Number of moles = Mass / Molar mass = 333 g / 28.02 g/mol ≈ 11.88 mol
2. Determine the number of moles of hydrogen:
Since we have an excess amount of hydrogen, we need to find how many moles are required to react completely with 11.88 moles of nitrogen.
From the balanced chemical equation for the formation of ammonia (NH₃) from nitrogen and hydrogen:
N₂ + 3H₂ → 2NH₃
We can see that 1 mole of nitrogen reacts with 3 moles of hydrogen.
Number of moles of hydrogen = 11.88 mol * (3 mol of H₂ / 1 mol of N₂) ≈ 35.64 mol
3. Determine the limiting reagent:
To determine the limiting reagent, we compare the number of moles of each reactant to the stoichiometric ratio in the balanced equation.
The stoichiometric ratio is 1 mole of N₂ to 3 moles of H₂.
Since we have 35.64 moles of H₂ and only 11.88 moles of N₂, N₂ is the limiting reagent.
4. Calculate the number of moles of ammonia formed:
From the balanced equation, we can see that 1 mole of nitrogen reacts to form 2 moles of ammonia.
Number of moles of ammonia formed = 11.88 mol * (2 mol of NH₃ / 1 mol of N₂) ≈ 23.76 mol
5. Convert moles of ammonia to volume at the given conditions:
Using the ideal gas law equation PV = nRT, we can rearrange it to find the volume:
V = (nRT) / P
Where:
V = Volume of gas (in liters)
n = Number of moles
R = Ideal gas constant (0.0821 L·atm/mol·K)
T = Temperature in Kelvin
P = Pressure in atmospheres
T = 235°C + 273.15 = 508.15 K
Now, we can calculate the volume of ammonia using the given pressure and temperature:
V = (23.76 mol * 0.0821 L·atm/mol·K * 508.15 K) / 7.0 atm
V ≈ 921.63 L
Therefore, the volume of ammonia formed is approximately 921.63 liters at 7.0 atm and 235°C.
Moving on to the second part of your question:
To calculate the density of hydrogen at STP (Standard Temperature and Pressure), we need to know the molar mass of hydrogen and the molar volume of a gas at STP.
The molar mass of hydrogen (H₂) is approximately 2.02 g/mol.
At STP, the temperature (T) is 273.15 K and the pressure (P) is 1.00 atm.
The molar volume at STP is 22.4 L/mol.
To calculate the density, we use the formula:
Density = Mass / Volume
From the molar mass of hydrogen, we know that the mass of 1 mole of hydrogen is 2.02 g.
Using Avogadro's Law, 1 mole of any gas occupies 22.4 L at STP.
Therefore, the density of hydrogen at STP is:
Density = 2.02 g / 22.4 L ≈ 0.090 g/L
So, the density of hydrogen at STP is approximately 0.090 g/L.