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 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