Answer:
Hydrogen Hâ‚‚ will be the limiting reagent.
The excess reactant that will be left after the reaction is 3.45 moles.
4.3 moles of water can be produced.
Explanation:
The balanced reation is:
2 H₂ + O₂ → 2 H₂O
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of each compound participate in the reaction:
To determine the limiting reagent, you can use a simple rule of three as follows: if by stoichiometry 1 mole of Oâ‚‚ reacts with 2 moles of Hâ‚‚, how much moles of Hâ‚‚ will be needed if 5.6 moles of Oâ‚‚ react?
[tex]moles of H_{2} =\frac{5.6 moles of O_{2} *2 mole of H_{2} }{1 mole of O_{2}}[/tex]
moles of Hâ‚‚= 11.2 moles
But 11.2 moles of Hâ‚‚ are not available, 4.3 moles are available. Since you have less moles than you need to react with 5.6 moles of Oâ‚‚, hydrogen Hâ‚‚ will be the limiting reagent and oxygen Oâ‚‚ will be the excess reagent.
Then you can apply the following rules of three:
[tex]moles of O_{2} =\frac{1 mole of O_{2} *4.3 mole of H_{2} }{2 mole of O_{2}}[/tex]
moles of Oâ‚‚= 2.15 moles
The excess reactant that will be left after the reaction can be calculated as:
5.6 moles - 2.15 moles= 3.45 moles
The excess reactant that will be left after the reaction is 3.45 moles.
[tex]moles of H_{2}O =\frac{2 moles of H_{2}O *4.3 mole of H_{2} }{2 mole of H_{2}}[/tex]
moles of Hâ‚‚O= 4.3 moles
4.3 moles of water can be produced.
