Nitrogen reacts with powdered aluminum according to the reaction:
2Al(s)+N2(g)→2AlN(s)
How many liters of N2 gas, measured at 892 torr and 98 ∘C, are required to completely react with 17.8 g of Al?
mols Al = grams/atomic mass = ?
Using the coefficients in the balanced equation, convert mols Al to mols N2. That is mols Al x (1 mol N2/2 mols Al) = ?
Now use PV = nRT with the conditions listed to solve for V. Post your work if you get stuck.
is this the right setup? (1.174) x V= (36)(0.0821)(371)
correction: (1.174) x V= (8.9)(0.0821)(371)
The set up is OK but n doesn't look right to me.
How do you calculate n specifically?
To find the number of liters of N2 gas required to react with 17.8 g of Al, we need to use the ideal gas law and stoichiometry.
First, let's convert the given temperature of 98 °C to Kelvin by adding 273:
98 °C + 273 = 371 K
Next, let's calculate the pressure in atm using the given pressure of 892 torr:
892 torr / 760 torr/atm = 1.1737 atm
Now, we need to convert the mass of Al to moles. The molar mass of Al is 26.98 g/mol:
17.8 g / 26.98 g/mol = 0.6593 mol
According to the balanced equation, 2 moles of Al react with 1 mole of N2. Therefore, 0.6593 mol Al will react with:
0.6593 mol Al × (1 mol N2 / 2 mol Al) = 0.3296 mol N2
Now, let's put all the values into the ideal gas law equation: PV = nRT
P = pressure in atm = 1.1737 atm
V = volume in liters (what we want to find)
n = number of moles of gas = 0.3296 mol
R = ideal gas constant = 0.0821 L·atm/(mol·K)
T = temperature in Kelvin = 371 K
Solving for V, we get:
V = nRT / P = (0.3296 mol) × (0.0821 L·atm/(mol·K)) × (371 K) / (1.1737 atm) = 7.423 L
Therefore, approximately 7.423 liters of N2 gas, measured at 892 torr and 98 °C, are required to completely react with 17.8 g of Al.