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Licemer1
1 month ago
15

What mass of carbon dioxide (co2) can be produced from 86.17 grams of c6h14 and excess oxygen?

Chemistry
1 answer:
lorasvet [2.7K]1 month ago
3 0
2C6H14 + 13O2 ---> 6CO2 +14H2O

Calculating the molar mass of C6H14: M(C6H14)=12.011*6 +1.008*14 ≈ 86.17 g/mol

Thus, 86.17 g of C6H14 corresponds to 1 mole.

                                  2C6H14 + 13O2 ---> 6CO2 +14H2O
based on the equation        2 mol                            6 mol
according to the question    1 mol                            3 mol

To determine M(CO2): M(CO2)= 12.011 + 2*15.999= 44.009 g/mol
Therefore, 3 mol CO2*44.009 g/1 mol CO2 ≈ 132.0 g CO2
Final answer: 132.0 g CO2


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To calculate the percent yield:

Percent yield = 100% × (Actual yield)/(Theoretical yield)

To determine the theoretical yield:

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We convert 5.29 g of NaBr to moles.

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Molar Mass of NaBr = 102.894 g/mol

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According to the stoichiometry of the reaction:

2 moles of NaBr yield 1 mole of Br₂

Thus, 0.05141 mole of NaBr will produce (0.05141×1/2) mole of Br₂, which is 0.0257 mole of Br₂

Theoretical yield = Expected mass of Br₂ from the reaction

= (Number of moles) × (Molar mass)

Molar mass of Br₂ = 159.808 g/mol

Theoretical yield of Br₂ = 0.0257 × 159.808 = 4.108 g

Calculating the percent yield:

Percent yield = 100% × (Actual yield)/(Theoretical yield)

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Hope this is helpful!!!

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