Asked by Mary
Oxygen vapor combines with hydrogen gas to produce liquid water. If 50.0 g of oxygen combines with 105 L of Hydrogen gas (at STP) how many grams of water could be produced?
Answers
Answered by
DrBob222
2H2 + O2 ==> 2H2O
mols O2 = grams/molar mass = estimated 1.5.
mols H2 = L x (1 mol/22.4 L) = estimated 4.7
Convert mols O2 to mols H2O.
That's 1.5 mols O2 x (2 mol H2O/1 mol O2) = 1.5 x 2/1 = estimated 3 mols H2O.
Convert 4.7 mols H2 to mols H2O.
That's 4.7 mols H2 x (2 mol H2O/2 mol H2) = 4.7 x 2.2 = estimated 4.7 mols H2O.
You have two different answers for mols H2O which means oe of them is wrong. The correct value in limiting reagent problems is ALWAYS the smaller value; therefore, 3 mol(estimated) H2O is formed.
Now convert mols H2O to grams.
g = mols x molar mass.
mols O2 = grams/molar mass = estimated 1.5.
mols H2 = L x (1 mol/22.4 L) = estimated 4.7
Convert mols O2 to mols H2O.
That's 1.5 mols O2 x (2 mol H2O/1 mol O2) = 1.5 x 2/1 = estimated 3 mols H2O.
Convert 4.7 mols H2 to mols H2O.
That's 4.7 mols H2 x (2 mol H2O/2 mol H2) = 4.7 x 2.2 = estimated 4.7 mols H2O.
You have two different answers for mols H2O which means oe of them is wrong. The correct value in limiting reagent problems is ALWAYS the smaller value; therefore, 3 mol(estimated) H2O is formed.
Now convert mols H2O to grams.
g = mols x molar mass.
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