Answer:
84 grams H2O is formed based on oxygen's status as limiting reagent, 31.5 grams of excess H2 is also remaining.
Step-by-step explanation:
The first thing we need is a balanced equation.
2H2 + O2 = 2H2O + [lots of noise and light (exothermic reaction)]
Note that we should expect 2 moles of water for every 2 moles of H2, a 1:1 molar ratio.
For O2, we should expect 2 moles of water for every 1 mole of O2, a 1:2 molar ratio.
Since we are given masses for both reactants, the question of limited reagent must be addressed. If one of the reactants in short supply, then it becomes the limiting regent that defines just how much H2) will be produced.
Next, lets convert the masses into moles. This will allow us to compare reagent amounts and whether one is limiting.
Moles
Reagent Mass(g) Molar Mass Moles(mass/molar mass)
H2 32 2 16
O2 84 32 2.63
We know from the balanced equation that 2 moles of H2 is needed for every 1 mole of O2. There are only 2.63 moles of O2, which when doubled becomes 5.25 moles. This is the number of moles of H2 that would be consumed if all of the O2 reacts. O2 is therefore the limiting reagent. Since there 16 moles of H2 to begin with, there will be (16 - 5.25) 15.75 moles of excess H2.
Knowing that O2 is the limited reagent allows us to determine how much H2O could be produced if all of the O2 were consumed.
The balanced equation says that we should get 2 moles of H2O for every 1 mole of O2. Therefore:
(2.63 moles O2)*(2 moles H2O/mole O2) = 5.25 moles H2O
Convert this to mass (g) by multiplying by the molar mass of water:
(5.25 moles H2O)*(18 g/mole H2O) = 84 grams
Remaining H2 is: (15.75 moles excess H2)*(2 g/mole H2) = 31.5 grams of excess H2