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?
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.
To determine the number of grams of water that can be produced, we need to calculate the stoichiometric ratio between oxygen and water.
The balanced chemical equation for this reaction is: 2H2(g) + O2(g) → 2H2O(l)
From the balanced equation, we can see that two moles of water are produced for every one mole of oxygen. First, we need to convert the mass of oxygen to moles.
The molar mass of oxygen (O2) is 32.00 g/mol, so we can calculate the number of moles of oxygen (n) using the formula:
n = mass / molar mass
n = 50.0 g / 32.00 g/mol ≈ 1.5625 mol
Since two moles of water are produced for every one mole of oxygen, we can use the stoichiometric ratio to find the number of moles of water produced. In this case, we have 1.5625 moles of oxygen, so the number of moles of water (n_water) will be:
n_water = 2 * n_oxygen
n_water = 2 * 1.5625 mol ≈ 3.125 mol
Finally, we can convert the moles of water to grams using the molar mass of water (H2O), which is 18.02 g/mol:
mass_water = n_water * molar mass_water
mass_water = 3.125 mol * 18.02 g/mol ≈ 56.31 g
Therefore, approximately 56.31 grams of water could be produced when 50.0 grams of oxygen combines with 105 liters of hydrogen gas at STP.