4 NH3(g) + 5 O2(g) --> 4 NO(g) + 6 H2O(g)
4 NH3(g): 4 moles @ 17.03 g/mole = 68.12 g NH3
5 O2(g) : 5 moles @ 32 g.mole = 160 g O2
we needed over twice the # of grams of O2 , as compared to NH3,...
they didn't add that,...
O2 is the limiting reagent
find the # of grams of NH3 that react:
4.75 g O2 @ 68.12 g NH3 / 160 g O2 = 2.02 g of NH3 actually do react
find How many grams of the excess reactant remains :
3.30 - 2.02 =
your answer: 1.28 grams of NH3 remain
4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g)
In a certain experiment, 3.30 g of NH3 reacts with 4.75 g of O2.
3.56 grams of NO form
