Question

Nitrogen reacts with powdered aluminum according to the reaction:
2Al(s)+N2(g)→2AlN(s)
Part A
How many liters of N2 gas, measured at 899 torr and 92 ∘C, are required to completely react with 20.5 g of Al?

Answer 1

(I'm very rusty on this but)

first you would convert g of Al to moles, conduct the dimensional analysis calculations as usual to find the moles of N2 required

then, use PV = nRT to find the volume of nitrogen required

got to get some errands done, I'll check on this asap

Answer 2

made any progress on the calculations so far?

Answer 3

Can you walk me through this?
I just know that the moles is I think 0.7598.

Answer 4

moles of Al is good so far

since each 2 mole of Al requires 1 mol N2, we would divide that by 2 and then substitute it for n into PV = nRT

PV = nRT where

P = pressure (given)
V = volume (the unknown)
R = the ideal gas constant (I couldn't find a conversion factor for Torr Celcius, I could only find one for Torr Kelvin, so you will use 62.363577 as your R)
T = temperature ** since our R value was in Kelvin you will need to convert Celsius to K by adding 273.15 to the celcius value

plug these into the equation and solve for V

Answer 5

Oh, no, I got a negative answer.
What is the actual answer?

Answer 6

@Vocaloid

Answer 7

PV = nRT

(899 torr)V = (0.7598/2)(62.363577)(92 + 173.15) solve for V