Final answer:
To calculate the grams of H2O produced from 12.0 g H2 and 34.0 g O2, you can use the concept of limiting reagents. The limiting reagent is the reactant that will be completely consumed. In this case, the moles of O2 are calculated to be 1.06 mol, which is less than the 2 moles required to completely react with 12.0 g of H2. Therefore, the amount of H2O produced is determined by the moles of O2, resulting in 38.17 g of H2O.
Step-by-step explanation:
To calculate the grams of H2O produced from 12.0 g H2 and 34.0 g O2, you need to first find the limiting reagent, which is the reactant that will be completely consumed. Let's calculate the number of moles for both reactants:
Moles of H2 = mass / atomic weight = 12.0 g / 2.02 g/mol = 5.94 mol
Moles of O2 = mass / atomic weight = 34.0 g / 32.0 g/mol = 1.06 mol
The balanced equation tells us that 2 moles of H2 produce 2 moles of H2O and that 1 mole of O2 will produce 2 moles of H2O. Since there is less than 2 moles of O2, it is the limiting reagent. Therefore, the amount of H2O produced will be determined by the moles of O2. Let's calculate the moles of H2O produced:
Moles of H2O = moles of O2 * (2 moles H2O / 1 mole O2) = 1.06 mol * (2 mol H2O / 1 mol O2) = 2.12 mol H2O
To convert moles of H2O to grams, use the formula:
Mass of H2O = moles of H2O * molecular weight = 2.12 mol * 18.02 g/mol = 38.17 g H2O
Therefore, 38.17 grams of H2O can be produced from 12.0 grams of H2 and 34.0 grams of O2.