[Cheistrynoob768]

Ammonia is produced from its elements in a formation reaction. If 8.90 × 10^24 molecules of ammonia are produced, how many molecules of hydrogen were used? 

My Attempt:

H molarmass = 1.01g/mol

NH3 = 8.90 x 10^24

n = m/M

8.90*10^24 = m / 1.01 g/mol

I couldn't finish it because I didn't know what to do next...

[UG]

All you need here is a balanced equation to find out how many molecules of hydrogen you would need.

[Borek]

What you did so far doesn't make much sense. You need to convert number of molecules of ammonia to number of moles, then follow as UG suggested.

[Big-Daddy]

What you did so far doesn't make much sense. You need to convert number of molecules of ammonia to number of moles, then follow as UG suggested.

I don't think you even need to convert to number of moles. n[NH3]=molecules[NH3]/N_A, n[H2]=(stoichiometric factor)*n[NH3], molecules[H2]=n[H2]*N_A so molecules[H2]=(stoichiometric factor)*n[NH3]*N_A=(stoichiometric factor)*molecules[NH3]*N_A/N_A=(stoichiometric factor)*molecules[NH3].

Mostly unnecessary algebra but I wanted to prove that for myself. (stoichiometric factor) has been kept in this form so as not to reveal the final solution to OP.

Therefore the only thing we need to do is balance the equation, N_A cancels out. This is pretty intuitively obvious too. 

[Borek]

Yes, you are right. I misread the problem and after seeing OP struggling with mass calculation I thought problem asks for mass of hydrogen.

[curiouscat]

*3 /2