# In the explosion of a hydrogen-filled balloon, 0.10g of hydrogen reacted with 0.80g of oxygen. How many grams of water vapor are formed? (Water vapor is the only product.) How do i go about solving...

In the explosion of a hydrogen-filled balloon, 0.10g of hydrogen reacted with 0.80g of oxygen. How many grams of water vapor are formed? (Water vapor is the only product.) How do i go about solving this? Thanks!

llltkl | Student

Gaseous hydrogen reacts with oxygen to produce water according to the equation:

`2H_2+O_2 rarr 2H_2O`

The number of moles of the reactants are:

H2->0.1/2 =0.05

O2->0.8/32=0.025

according to the reaction stoichiometry,

2 moles of hydrogen require 1 mole of oxygen

so, 0.05 moles of hydrogen require 1*0.05/2, i.e. 0.025 moles of oxygen.

Exactly 0.025 moles of oxygen is present in the reaction mixture.

So, both the reactants would react and exhaust each other out.

From the balanced equation,

2 moles of hydrogen reacts with sufficient oxygen to produce (2*+16), i.e. 18 g H2O

Therefore, 0.05 moles of hydrogen would produce 18*0.05/2, i.e. 0.45 g H2O.

ayl0124 | Student

`2"H_2" + "O_2" -> "H_2O"`

First, you must find the limiting reagent. For every mole of O2 used, two moles of H2 is needed.

`0.10"g(H_2)" = (1"mol(H_2)")/(2.016"g(H_2)") = 0.050"mol(H_2)"`

`0.80"g(O_2)" = (1"mol(O_2)")/(32"g(O_2)") = 0.025"mol(O_2)"`

Neither is the limiting reagent in this problem (for 0.025 moles of O2, you need 0.50 moles of H2). You can use either value for stoichiometry.

`0.025"mol(O_2)" = (1"mol(H_2O)")/(1"mol(O_2)") = (18.016"g(H_2O)")/(1"mol(H_2O)") = 0.4502"g(H_2O)"`

You will produce 45 grams of H2O. There needs to be two sig figs in your final answer.