Oxygen gas may be prepared by decomposing hydrogen peroxide, h2o2, using manganese (IV) oxide catalyst: 2H2O2(liquid) ----> 2H2O(liquid) + O2(gas)
Calculate the amount of oxygen that will be produced from 1.35 moles of H2O2
Determine the mass of oxygen gas that can be prepared from 2.46 g of hydrogen peroxide
To calculate the amount of oxygen gas produced from a given amount of hydrogen peroxide, we need to use the stoichiometry of the reaction.
1.35 moles of H2O2 are given. By examining the balanced chemical equation, we can see that for every 2 moles of H2O2, 1 mole of O2 is produced.
So, we can set up a ratio:
(1.35 mol H2O2) x (1 mol O2 / 2 mol H2O2) = 0.675 mol O2
Therefore, 1.35 moles of H2O2 will produce 0.675 moles of O2.
To determine the mass of oxygen gas produced from 2.46 grams of hydrogen peroxide, we need to convert grams to moles first.
The molar mass of H2O2 is 34.02 g/mol.
So, we have:
2.46 g H2O2 x (1 mol H2O2 / 34.02 g H2O2) = 0.0723 mol H2O2
Now, using the stoichiometry of the reaction as before, we can calculate the moles of O2 produced:
(0.0723 mol H2O2) x (1 mol O2 / 2 mol H2O2) = 0.03615 mol O2
Finally, to determine the mass of O2, we multiply the number of moles by the molar mass of O2, which is 32.00 g/mol:
0.03615 mol O2 x 32.00 g/mol = 1.1568 g O2
Therefore, 2.46 grams of H2O2 will produce about 1.16 grams of O2.