N2O4 and N2H4 react with each other to form NO and H2O. What is the maximum mass (in grams) of NO that could be obtained from 15.5 g of N2O4 and 4.68 g of N2H4 when they react?
The balanced chemical equation for the reaction is:
Molar mass of `N_2H_4=14*2+1*4=32` g/mol
Molar mass of `N_2O_4=14*2+16*4=92` g/mol
15.5 g `N_2O_4` =`15.5/92` =`0.168478` moles
4.68 g `N_2H_4` =`4.68/32` =`0.14625` moles
1 mol `N_2H_4` reacts with 2 mol `N_2O_4`
Hence, 0.14625 moles `N_2H_4` require 2*0.168478=0.336956 moles `N_2O_4` for the reaction
But there is only 0.168478 moles of `N_2O_4` .
Hence, `N_2O_4` is the limiting reagent.
From the stoichiometry of the given reaction, considering the limiting reagent:
2 mol `N_2O_4` reacts with sufficient `N_2H_4` to produce `6*30=180` g `NO` .
Hence, 0.168478 moles of `N_2O_4` reacts with sufficient `N_2H_4` to produce `(180*0.168478)/2=15.16302~~15.2` g `NO` .
Therefore, 15.2 g `NO` can be obtained from 15.5 g of `N_2O_4` and 4.68 g of `N_2H_4` when they react.