How many grams of the excess reactant remain after the limiting reactant is completely consumed? express your answer using two significant figures?

 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 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 

You know how the first one if filled up like the objects do the same but with the other categories