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The equation below shows the incomplete combustion of ethene.

C2H4 + O2 → CO2 + H2O

If 2.70 mol C2H4 is reacted with 6.30 mol O2 identify the limiting reagent

Respuesta :

Edufirst

Answer:

  • The limiting reagent is O₂.

Explanation:

1. Write the balanced molecular equation:

     [tex]C_2H_4+3O_2\longrightarrow 2CO_2+2H_2O[/tex]

2. Write the mole ratio of the reactants

The theoretical mole ratio is given by the coefficients of the balanced molecula equation: 3 moles of O₂ react with 1 mol of C₂H₄

       [tex]\dfrac{3molO_2}{1molC_2H_4}[/tex]

3. Write the mole ratio of the substances available

     

       [tex]\dfrac{6.3molO_2}{2.70molC_2H_4}{\approx \dfrac{2.3molO_2}{1molC_2H_4}[/tex]

4. Conclusion

Since the ratio of the O₂ to C₂H₄ available is less than the theoretical ratio of O₂ needed to react with C₂H₄, the former is not enough: it will be consumed before the other reagent and it is the limiting reactant.

Hence, the limiting reagent is O₂.

The limiting reagent is O₂.

Firstly, write the balanced chemical equation:

C₂H₄ + O₂ → CO₂ + H₂O

Then, we need to write the mole ratio of the reactants

The theoretical mole ratio is given by the coefficients of the balanced chemical equation: 3 moles of O₂ react with 1 mol of C₂H₄

[tex]\frac{\text{3 mol O_2}}{\text{1 mol C_2H_4}}[/tex][tex]\frac{3 mol O_2}{1 mol C_2H_4}[/tex]        

Now, let's write the mole ratio of the substances available

[tex]\frac{6.3 mol O_2}{2.70mol C_2H_4}[/tex]  ≈   [tex]\frac{2.3 mol O_2}{1 mol C_2H_4}[/tex]

We concluded that,

Since, the ratio of the O₂ to C₂H₄ available is less than the theoretical ratio of O₂ needed to react with C₂H₄, the former is not enough: it will be consumed before the other reagent and it is the limiting reactant.

Hence, the limiting reagent is O₂.

Learn more:

brainly.com/question/18803167

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