Taking into account definition of percent yield, the amount of water produced is 32.4 grams.
In first place, the balanced reaction is:
2 C₂H₂ + 5 O₂ → 4 CO₂ + H₂O
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.
To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 5 moles of O₂ reacts with 2 moles of C₂H₂, 14.8 moles of O₂ reacts with how many moles of C₂H₂?
[tex]moles of C_{2} H_{2} =\frac{14.8 moles of O_{2}x 2 moles of C_{2} H_{2}}{5 moles of O_{2}}[/tex]
moles of C₂H₂= 5.92 moles
But 5.92 moles of C₂H₂ are not available, 4.8 moles are available. Since you have less moles than you need to react with 14.8 moles of O₂, C₂H₂ will be the limiting reagent.
The percent yield is the ratio of the actual return to the theoretical return expressed as a percentage and it is calculated as the experimental yield divided by the theoretical yield multiplied by 100%:
percent yield= (actual yield÷ theorical yield)× 100%
where the theoretical yield is the amount of product acquired through the complete conversion of all reagents in the final product, that is, it is the maximum amount of product that could be formed from the given amounts of reagents.
Considering the limiting reagent, the following rule of three can be applied: if by reaction stoichiometry 2 moles of C₂H₂ form 1 mole of H₂O, 4.8 moles of C₂H₂ form how many moles of H₂O?
[tex]moles of H_{2} O=\frac{4.8 moles of C_{2} H_{2} x1 mole ofH_{2} O }{2moles of C_{2} H_{2}}[/tex]
moles of H₂O= 2.4 moles
The molar mass of water is 18 g/mole. Then, the theorical mass of water formed can be calculated as: 2.4 moles ×18 g/mole= 43.2 grams
In this case, you know:
Replacing in the definition of percent yields:
75%= (actual yield÷ 43.2 g)× 100%
Solving:
75%÷100%= actual yield÷ 43.2 g
0.75= actual yield÷ 43.2 g
0.75× 43.2 g= actual yield
32.4 g= actual yield
In summary, the amount of water produced is 32.4 grams.
Learn more about
the reaction stoichiometry:
brainly.com/question/24741074
brainly.com/question/24653699
percent yield:
brainly.com/question/14408642
#SPJ1
