Need help with my last question!

Question

Given the following reaction:

2NH3 (g) + 5F2 (g) ---> N2F4 (g) + 6HF (g)

If 4.00 g of NH3 and 14.0 g of F2 are allowed to react, how many grams of HF can be produced?

Answer 1

grams/mol
N=14
H=1
F= 19
so
NH3 =17
F2 = 38
HF = 20
so
If I have 4 g of NH3, how many g of F2 do I need?
4/[2(17)] = x/[5(38)]
x = 20*38/34 = 22.4 g F2

whoops, not enough F2

so use 14 of F2 and y of NH3
y/34 = 14/190
y = 2.5 grams of NH3

then
2.5/34 = HF/[6(20)]
HF = 8.82 grams HF

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check that should be 14 grams of F2
14/[5(38)] = 8.8/[6(20)]
.074 = .073 close enough