Help with chemistry please?

Question

I am stuck on a question it says:
Calculate the mass of nitrogen required to produce 250dm3 (decimeters cubed) of ammonia at 723K and 2.00 x 10^7 Pa. (the volume of 1 mole of any gas at 298K and 1.01 x 10^5 Pa is 24 dm3).
Please help! Do not waste my time by making smart comments like do it yourself you'll just be reported... Many thanks to those who answer seriously!!! :-)

Answer 1

you'll have to use PV=nRT to find n (number of moles), then use that number to convert to grams (which is just dividing by the grams/mole of Nitrogen, 14).

So n = PV/RT. You know what R is, it's always 0.08206 (L atm / mol K). You'll need to do a lot of converting so that the units cancel out.

Let's start with volume. You have 250 dm^3.
10^10 dm = 1 m and 1m = 100 cm
SO:
(250 dm^3) x [1m / (10^10 dm)]^3 x [100cm / 1m]^3 = X (cm^3)

Why did we do this? because 1cm^3 = 1mL and 1000mL = 1L.
So divide the your X by 1000 and that's your volume V.

Next is to convert your pressure.
1 Pa = 9.86923267 × 10^-6 atm

So (2.0x10^7Pa) x [(10^-6 atm) / (1 Pa)] = P, your pressure in atmospheres.

Your temp is already in Kelvin, so we don't have to convert.

Now, we're ready to solve for n, your number of moles:
n = PV / RT

Once you find n, which is in units of moles, we know that there is 1 mol of N in 14 grams of N. Therefore:

n (mol) x [14 g / 1 mol] = YOUR FINAL ANSWER!

Yay, we solved it. I would help you with the numbers but I don't have a calculator with me.

Keep going at it. You'll be thankful if you know your PV=nRT's very well later on.

Answer 2

First of all we need to know the chemical reaction that produces the ammonia from gaseous nitrogen:

N2 + 3H2 ----> 2 NH3

It means that 1 mole of nitrogen yields 2 moles of Ammonia

1 mole of N2 = 14 x 2 = 28 g/mole
1 mole of NH3 = 14 + 3 = 17 g/mole

So, 28 g of N2 yields 17 x 2 = 34 g of ammonia......(1)

As we need to know the mass of Nitrogen that produces the given volume of ammonia, let's check what is the mass of 250 dm³ of that gas.

Remember that Ideal gas law tell us:

PV = nRT

where : P= pressure of gas
..............V = Volume of gas
..............n = number of moles
..............R = gas constant
..............T = absolute temperature

On the other hand we know that n= m/M

where: m = mass
............M = molecular weight

so, the ideal gas law becomes:

PV = (m/M)RT we get m as follows:

m = PVM/RT

where : P=2x10^7 Pa
.............V = 250 dm³ = 0.250 m³
.............M (NH3) = 17 g/mole = 0.017 Kg/mole
.............T = 723 K

m = (2 x 10^7 Pa)(0.250m³)(17 g/mole) / (8.31 J K^-1 mol^-1)(723 K)
m = 14147.5 g of ammonia

Going back to our chemical equation:

How many grames of nitrogen yields 14147.5 grames of ammonia?

In row (1) we said that: 28 g of N2 produce 34 g of NH3
.................................so: x g of N2 produce 14147.5 g of NH3

x = 11650.08 g of Nitrogen

That´s it!

I am writing you from Mexico so excuse some miswriting that I could have.

Good luck! (!Buena suerte!)

Answer 3

Use PV=nRT to calculate the number of moles of AMMONIA:

n=PV/RT

R for the same system of units you are using is

8.314472 (Pa x m^3) / (mol x K)

P and T are in the correct units but V needs to be changed:

1 m^3 = 1000 dm^3 ===> 250 dm^3 = 0.25 m^3

replace the values in the above equation and you get:

n= 831.75 mol of NH3

You are being asked for mass of NITROGEN not ammonia. Look at the formula of ammonia: 1 mol of ammonia contains one mole of N ATOMS, thus:

831.75 mols of NH3 contain 831.75 mols of N atoms

once you have the mols of N we need to multiply the atomic weight of N to get the mass:

831.75 mol x 14.0 g/ mol = 11644.5 g of nitrogen.

Answer 4

1st use the combined gas law to convert your given volume to the volume at STP.
The convert the new volume into moles by dividing by 22.4
Lastly, convert your moles to mass by multiplying by the molecular mass of Nitrogen.