[stoicheomexic]

Just made this account , because i want to be sure if i solved this exercices correct
Would really appreciate it if somebody could just point me in the right direction if i got any wrong
N2 + 3 H2 = 2 NH3
N2 = 125 grams 4.5 moles
H2 = 32 grams 48 moles
NH3 = 36.5 grams 4.2 moles
Rendement (=output according online translator)(NH3) = 51% >>correct??

2 C6H6 + 15 O2 = 12 CO2+ 6 H2O
C6H6 = 10 grams
H2O = 6.9 grams >> correct??
Volume CO2 = 103 liters >>correct??

C2H6 = C2H4 + H2
C2H4 = 30 liters
rendement (C2H4) = 67% >> correct?

[Dan]

N2 + 3 H2 = 2 NH3
N2 = 125 grams 4.5 moles
H2 = 32 grams 48 moles
NH3 = 36.5 grams 4.2 moles
Rendement (=output according online translator)(NH3) = 51% >>correct??

The highlighted mole calculations seem to have gone wrong, and consequently your yield (rendement = yield) is wrong.

2 C6H6 + 15 O2 = 12 CO2+ 6 H2O
C6H6 = 10 grams
H2O = 6.9 grams >> correct??
Volume CO2 = 103 liters >>correct??

I agree with your mass of water, but not with your volume of CO₂, can you show working?

C2H6 = C2H4 + H2
C2H4 = 30 liters
rendement (C2H4) = 67% >> correct?

This calculation is impossible without knowing how much ethane you started with...