Answer:
94.57 g H2O
Explanation:
32.0 g H2
Molar Mass H2: 2.01 g/mol
84.0 g O2
Molar Mass O2: 32.00 g/mol
Molar Mass H2O: 18.02 g/mol
First balance the eqution:
2H2 + O2 = 2H2O
Find the amount of moles that both 32.0 g of H2 and 84.0 g of O2 can produce using the molar masses:
32.0 g H2 x ((1 mole H2)/ (2.015 g H2)) = 15.88 moles H2
84.0 g O2 x ((1 mole O2)/(32.00 g O2)) = 2.63 moles O2
Now you can find the amount of grams of H2O each reactant will produce:
**keep in mind of mole ratios from balanced equation**
15.88 moles H2 x ((2 moles H2O)/(2 moles H2)) x ((18.015 g H2O)/(1 mole H2O) =286.07 g H2O
2.63 moles O2 x ((2 moles H2O)/(1 mole O2)) x ((18.015 g H2O)/(1 mole H2O) = 94.76 g H2O
94.76 is the final answer because this is the limiting reactant, meaning it produces less product than the H2 so it limits the reaction from producing anymore product from the amount calculated.
