Answer:
11.28 grams of H2O can be produced from 10.0 grams of O2 with excess H2.
Step-by-step explanation:
The balanced chemical equation for the reaction between oxygen and hydrogen to form water is:
2 H2 + O2 → 2 H2O
According to the equation, 1 mole of O2 reacts with 2 moles of H2 to produce 2 moles of H2O. Therefore, we need to first calculate the number of moles of O2 we have and then use stoichiometry to determine the number of moles of H2O that can be produced.
Molar mass of O2 = 32.00 g/mol
Number of moles of O2 = mass / molar mass = 10.0 g / 32.00 g/mol = 0.3125 mol
From the balanced equation, we know that 1 mole of O2 reacts with 2 moles of H2 to produce 2 moles of H2O. So the number of moles of H2O that can be produced is:
0.3125 mol O2 × (2 mol H2O / 1 mol O2) = 0.625 mol H2O
Finally, we can calculate the mass of H2O produced using its molar mass:
Molar mass of H2O = 18.02 g/mol
Mass of H2O produced = number of moles × molar mass = 0.625 mol × 18.02 g/mol = 11.28 g