Consider the unbalanced equation for the oxidation of butene. C4H8 + 6O2 CO2 + H2O For each molecule of C4H8 that reacts, how many molecules of carbon dioxide and water are produced? two carbon dioxide molecules and two water molecules
four carbon dioxide molecules and four water molecules
two carbon dioxide molecules and four water molecules
four carbon dioxide molecules and two water molecules

Four carbon dioxide molecules and four water molecules

Explanation:

D: four carbon dioxide molecules and two water molecules

Explanation:

There are four carbon dioxide molecules (CO₂) on the right side of chemical reaction, because there are four carbon atoms on the left side of balanced chemical reaction.

There are four water molecules (H₂O) on the right side of chemical reaction, because there are eight hydrogen atoms on the left side of balanced chemical reaction.

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.

1.solid

2.carbon dioxide

3.helium