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# Help with moles calculation please!!! watch

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1. 2NH3 + NaOCl -> N2H4 + NaCl + H2O

Damn... it didn't copy right.. anyway all the numbers are supposed to be subscripts appart from the 2 in front of Ammonia.

What mass of ammonica is needed to make 64g of hydrazine (N2H4)
H = 1 and N = 14

I got 17g.. it seems wrong to me, can you help please??
Thanks!
2. I haven't done chemistry in ages.. but am I the only one who gets 68g?
3. I got 68g as well...
4. Moles of N2H4 = mass/ Mr = 64/(14*2 + 3) = 2 moles

Since the mole ratio of NH3 : N2H4 is 2 : 1
4 moles of NH3 would be required to form 2 moles of N2H4.

Mass of NH3 = moles * Mr = 4 * (14+3) = 68g
5. (Original post by xPunkx)
2NH3 + NaOCl -> N2H4 + NaCl + H2O

Damn... it didn't copy right.. anyway all the numbers are supposed to be subscripts appart from the 2 in front of Ammonia.

What mass of ammonica is needed to make 64g of hydrazine (N2H4)
H = 1 and N = 14

I got 17g.. it seems wrong to me, can you help please??
Thanks!
12st work out MR of the hydrazine: which is 32. Moles is equal to mass divided by Mr. thereforenumber of moles is 2. Moles ratio is 2:1 therfore you ahve a total of 4 moles of ammonia. Mr of ammonia is 17. times by 4 gives 68g.

someone bet me to it! lol!
6. Yay I've got it now. I'd got the ratio right but put it as 1 mole instead of 4 somehow =/ I'm not thinking straight. haha. Thanks for your help!
7. Phew, I haven't lost touch with the subject......... yet.
8. yep 68g

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