Question

2 C4H10 + 13O2---->8 CO2 + 10 H20

How many moles of water are produced when 1.4 moles of C4H10 react completely with excess oxygen?

If in the reaction described above produces 5 moles of water, what is the percent yeild?

If There is excess C4H10, HOw many grams of O2 are needed to produce 85 grams of CO2?

Answer 1

2 C4H10 + 13O2   --------------------->8 CO2 + 10 H20

from the balanced equation :

2 mol C4H10    -------------------> 10 mol water

1.40 mol C4H10   ---------------->   ??

moles of water produced = 1.4 x 10 / 2 = 7 mol

moles of water produced = 7.0 mol

2)

percent yield = (actual / theoretical ) x 100

                      = (5 / 7 ) x 100

percent yield = 71.4 %

3)

moles of CO2 = 85 / 44 = 1.932 mol

8 mol CO2   -----------------> 13 mol O2

1.932 mol   ----------------> ??

moles of O2 needed = 1.932 x 13 / 8 = 3.14 mol

moles of O2 needed = 3.14 mol

mass of O2 needed = 100.45 g