Chemistry help needed!!! Pls. quick>>>?

Question

N2 + 3H2------2NH3
If N2 is 2kg and H2 is 1kg how much NH3 will be evolved? The answer is 2.571 kg but can anyone pls. explain how you're getting it.

Answer 1

Assuming that the reaction only goes one way (which it doesnt but just pretend), you calculate the number of mols of each reactant on the left hand side of the equation. You do this by dividing the mass by the formula weight (get your periodic table)

2kg N2 = 2000g, N2 = (14+14) = 28g/mol

so 2000g divided by 28g/mol = 71.4 mol

for H2 = 1000g, H2 = 2g/mol

so 1000g divided by 2g/mol = 500mol

each N2 molecule reacts with 3 H2 molecules (from the equation) so 71.4 mols N2 requires 214.2 mols H2, so obviously the H2 is in great excess. In other words N2 is your limiting factor and you have to use it to calculate the amound of product formed.

For each mol of N2 you get 2 mols NH3, so if you start with 71.4 mols N2 you will get 2 x 71.4mols = 142.8mols NH3

multiply this value by the formula weight of NH3 (14+1+1+1) = 17g/mol

and you get 2427g... ok so this isnt exactly the answer you said but iv been rounding my numbers, so if you have ascientific calculator youll hopefully get a more precise answer. This is the correct method though.

hope it helps.

Answer 2

moles N2 = 2000 g / 28 g/mole = 71.4 moles
moles H2 = 1000 g / 2 g/mole = 500 moles

You need 3 times as many moles of H2 to react with the N2 completely. Since you have more than that, N2 is your limiting reactant. Since you have 71.4 moles N2, you can get 142.8 moles of NH3.

142.8 moles NH3 * 17 g/mole = 2427 g = 2.427 kg (rounding)