How many grams of H2O will be formed when 32.0 g H2 is mixed with 84.0 g of O2 and allowed to react to form water

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.