The percent yield of the reaction has been 74.47%.
The balanced chemical equation for the reaction has been:
[tex]\rm 2\;H_2\;+\;O_2\;\rightarrow\;2\;H_2O[/tex]
(a) The theoretical yield can be calculated as:
The theoretical yield of water has been:
1 mole of oxygen produces = 2 moles of water
0.1562 moles of oxygen produces = 0.1562 [tex]\times[/tex] 2
= 0.312 moles of water.
The mass of water has been = moles [tex]\times[/tex] molecular weight
The theoretical yield of water = 0.312 [tex]\times[/tex] 18 grams
The theoretical yield of water = 5.616 grams.
(b) The limiting reactant has been the one whose concentration has been in a lesser quantity for the complete utilization of other reactants.
According to the reaction, for the complete utilization of 2 moles of Hydrogen, 1 mole of oxygen has been used.
The moles of 5 grams Hydrogen:
Moles = [tex]\rm \dfrac{weight}{molecular\;weight}[/tex]
Moles of hydrogen = [tex]\rm \dfrac{5}{1}[/tex]
Moles of hydrogen = 5 mol.
Moles of Oxygen = [tex]\rm \dfrac{5}{32}[/tex]
Moles of oxygen = 0.1562 mol.
For the complete utilization of 5 moles of hydrogen 2.5 moles of oxygen has been required.
Since the oxygen has been present in a lesser quantity, the oxygen has been the limiting reactant.
(c) The consumption of 0.1562 moles of Oxygen has required 0.312 mol of Hydrogen. The hydrogen has been present in a higher quantity than required.
Thus. hydrogen has been the excess reactant.
(d) The remained excess reactant has been:
Remained = Total - consumed
Remained = 5 - 0.312
Remained = 4.68 mol.
The excess reactant has been 4.68 mol extra than required.
(e) The theoretical yield of water has been:
1 mole of oxygen produces = 2 moles of water
0.1562 moles of oxygen produces = 0.1562 [tex]\times[/tex] 2
= 0.312 moles of water.
The mass of water has been = moles [tex]\times[/tex] molecular weight
The theoretical yield of water = 0.312 [tex]\times[/tex] 18 grams
The theoretical yield of water = 5.616 grams.
The actual yield of water has been 4.2 grams.
The percent yield of water = [tex]\rm \dfrac{Actual\;yield}{Theoretical\;yield}\;\times\;100[/tex]
The percent yield of water = [tex]\rm \dfrac{4.2}{5.616}\;\times\;100[/tex]
The percent yield of water = 74.74 %
For more information about the limiting reactant, refer to the link:
https://brainly.com/question/14225536
