2H2 (g) + O2 (g)→ 2H2O (l)

Question

A total of 12.0 grams of hydrogen and 64.0 grams of oxygen are available. Once mixed, this reaction runs to completion, meaning that all of the limiting reactant is consumed.
a. Which reactant is limiting?
b. How many moles of water are formed?
c. How many moles of hydrogen are consumed?
d. How many moles of oxygen are consumed? e. What is the mass of oxygen that is consumed?
f. What is the mass of water that is formed?

Answer 1

2H2 (g) + O2 (g)→ 2H2O (l)
A total of 12.0 grams of hydrogen and 64.0 grams of oxygen are available. Once mixed, this reaction runs to completion, meaning that all of the limiting reactant is consumed.
a. mols O2 = grams/molar mass = 64.0/32 = 2
mols H2 = 12.0/2 = 6
mols H2O formed from 2 mols O2 and excess H2 = 4
mols H2O formed from 6 mols H2 and excess O2 = 6
In limiting reagent problems (LR) the smaller number always wins because you never can get more then the smallest possible. So O2 is the LR.
b. 2 mols O2 x (2 mol H2O/1 mol O2) = 4 mols H2O formed.
c. 2 mol O2 x (2 mols H2/1 mol O2) = 4 mols H2 used.
d. all O2 is consumed; i.e., 2 mols.
e. all of it is 64.0 g
f. mass water = mols H2O x molar mass H2O = ?