2 C2H2(g) + 5 O2(g) yields 4 CO2 (g) + H2O (g) How many grams of water can be produced by the reaction of 4.8 moles of C2 H2 with 14.8 moles of O2 if the percentage yield is 75% how many grams of water was produced

Respuesta :

Taking into account definition of percent yield, the amount of water produced is 32.4 grams.  

Reaction stoichiometry

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:

  • C₂H₂: 2 moles
  • O₂: 5 moles
  • CO₂: 4 moles
  • H₂O: 1 mole

Limiting reagent

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.

Limiting reagent in this case

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.

Percent yield

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.

Theorical mass of H₂O formed

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

Actual mass of H₂O formed

In this case, you know:

  • percent yield= 75%
  • actual yield= ?
  • theorical yield= 43.2 grams

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

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