Balanced equation first.
2H2 + O2 -> 2H2O
0.600 moles H2 (1 mole O2/2 moles H2)(6.022 X 10^23/1 mole O2)
= 1.81 X 10^23 molecules of O2 required
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Twelve, because each molecule of water has only one oxygen atom, while each mole of oxygen has two atoms of oxygen.
2H2 + O2 -> 2H2O
Therefore, 12.0 moles of water.
only 1 mole of water can be formed. Through stoichiometry you can find that hydrogen is the limiting reactant and it only produces 1 mol of water, or aprox. 18 grams of H2O.
5.40 mol H2O
3 moles of oxygen.
13 mole
No moles of oxygen are produced by complete combustion of propane. Oxygen is CONSUMED, not produced. For combustion of 4 moles of propane, it will use 20 moles of oxygen.
Only when 5.5 mole O2 react with 11 mole H2, then 11 mole H2O are formed.
6 moles COULD be produced
12 moles KClO3 (3 moles O/1 mole KClO3) = 36 moles of oxygen.
The reaction requires 2 moles of hydrogen gas and 1 mole of oxygen gas to produce 2 moles of water.
No moles of oxygen are produced by complete combustion of propane. Oxygen is CONSUMED, not produced. For combustion of 4 moles of propane, it will use 20 moles of oxygen.
Two moles of water are produced.
Only when 5.5 mole O2 react with 11 mole H2, then 11 mole H2O are formed.
6 moles
6 moles COULD be produced
12 moles KClO3 (3 moles O/1 mole KClO3) = 36 moles of oxygen.
The reaction requires 2 moles of hydrogen gas and 1 mole of oxygen gas to produce 2 moles of water.
2s + 3o2 --> 2so3
30 moles
10,55 moles of water are obtained.
2KClO3 --> 2KCl + 3O2For every 3 moles of oxygen gas produced, 2 moles of potassium chlorate are used.6 moles O2 * (2 moles KClO3 reacted / 3 moles O2 produced) = 4 moles KClO3
.913 moles