<u>Given information:</u>
Mass of H2 = 2 g
Mass of O2 = 32 g
<u>To determine:</u>
Mass of H2O2 produced
<u>Explanation:</u>
The reaction between H2 and O2 can be given as:
H2 + O2 → H2O2
Based on the reaction stoichiometry:
1 mole of H2 reacts with 1 mole of O2 to form 1 mole of H2O2
# moles of H2 = mass of H2 / molar mass of H2 = 2 g/ 2 g.mol-1 = 1 mole
# moles of O2 = mass of O2/ molar mass of O2 = 32 g/ 32 g.mol-1 = 1 mole
Hence for the given reactant conditions, moles of H2O2 produced = 1
Mass of H2O2 = moles of H2O2 * molar mass H2O2 = 1 mole * 34 g.mole-1 = 34 g
<u>Ans</u>: 34 g of H2O2 is produced in this reaction
Answer:
4.525% is the percentage by volume of oxygen in the gas mixture.
Explanation:
Total pressure of the mixture = p = 4.42 atm
Partial pressure of the oxygen = 
Partial pressure of the helium = 
(Dalton law of partial pressure)





According Avogadro law:
(At temperature and pressure)
Volume occupied by oxygen gas =
Total moles of gases = n = 1 mol
Total Volume of the gases = V


Percent by volume of oxygen in the gas mixture:

Your answer is the element of S I think.