[Lo138]

If you wanted to produce 15,000 kg of ammonium nitrate, what would the minimum volume of 12.0M nitric acid AND minimum volume of 8.00M aqueous ammonia that would be required?

Please Show All Work. Thanks in advance to anyone that answers or helps.

[Astrokel]

It is against the forum rule to do work for you. Show your attempt. Start off with balanced equation

[Lo138]

Well, starting off, I'm thinking I should convert from kilograms to grams to moles. And then I'm thinking I should use the M = m/v formula. Any help on what steps to take would be much appreciated, or at least, somewhere to start.

[Lo138]

I have completed the balanced equation as:

HNO3(aq) + NH3(aq) -> NH4NO3(aq)

[Astrokel]

You will have to form two equations and solve. I will start you off,

Let x and y (in L) be the volume of nitric acid and ammonia required respectively.

[Lo138]

NH3 + HNO3 ---> NH4NO3

This means 1 mol of NH3 and 1 mol of HNO3 is needed to make 1 mol of NH4NO3

Molar mass of NH4NO3
N = 14.01 g/mol*2
H = 1.008 g/mol*4
O = 16.00 g/mol*3

NH4NO3 = 80.052 g/mol

15,000 kg of anything is actuall 15,000,000 grams
So, 15,000,000/80 = 187,500 mol of NH4NO3

Means you need 187,500 mol of nitric acid, as well as 187,500 mol of ammonia.

First, nitric acid:
187,500 mol/12.0 mol/L = 15,625 L of nitric acid

And for ammonia:
187,500 mol/8 mol/L = 23,422 L of aqueous ammonia

[Lo138]

Can someone check my work and significant figures as well.

[Astrokel]

Correct, but your final answer should be 3 sf.