Which property of matter changes during a chemical change but not during a physical change

Final answer:

The balanced chemical equation for the combustion of octane is 2 C8H18 + 25 O2 → 16 CO2 + 18 H2O. From the stoichiometry of the reaction, 3.50 mol of CO2 will need approximately 50g of octane.

Explanation:

The combustion of octane (C8H18) in the presence of oxygen (O2) produces carbon dioxide (CO2) and water (H2O). The balanced chemical equation for this reaction is: 2 C8H18 + 25 O2 → 16 CO2 + 18 H2O.

To calculate the mass of octane needed to release 3.50 mol CO2, we need to understand the stoichiometry of the reaction. From the balanced equation, we can see that 16 mol of CO2 is produced from 2 mol of C8H18. So, 1 mol of C8H18 produces 8 mol of CO2.

Therefore, to produce 3.50 mol of CO2, we would need 3.50/8 = 0.4375 mol of C8H18. The molar mass of octane is approximately 114 g/mol, so the required mass would be 0.4375mol x 114g/mol = approximately 50g.

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Final answer:

The combustion of octane produces carbon dioxide and water, as described by the balanced chemical equation C8H18 + 12.5O2 → 8CO2 + 9H2O. To produce 3.50 mol of CO2, approximately 50g of octane is needed.

Explanation:

The combustion of octane (C8H18) in the presence of oxygen (O2) produces carbon dioxide (CO2) and water (H2O). The balanced chemical equation for this reaction is: C8H18 + 12.5O2 → 8CO2 + 9H2O.

Considering the stoichiometry of the reaction, we can see that 1 mol of octane produces 8 moles of CO2. Therefore, to produce 3.50 mol of CO2, the amount of octane required would be 3.50/8 = 0.4375 mol. Converting this to mass using the molar mass of octane (114.22 g/mol), we get 0.4375 mol * 114.22 g/mol ≈ 50 g. Thus, approximately 50g of octane is needed to produce 3.50 mol of CO2.

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Answer:

D) Life will no longer be possible

Explanation:

Odyssey ware

The percentage yield of the reaction has been 46.74%.

Percentage yield can be given as the ratio of the actual yield to the theoretical yield.

For the theoretical yield, the balanced equation will be:

Thus, 2 moles of ZnS will give 2 moles of ZnO.

The moles of 46.5 g ZnS:

Moles =

ZnS =

ZnS = 0.477 mol

The moles of 13.3 grams Oxygen:

Oxygen =

Oxygen = 0.415 mol.

The moles of 18.14 grams ZnO:

ZnO =

ZnO = 0.222 mol

According to the balanced chemical equation,

2 moles ZnO = 2 moles ZnS

0.222 mol ZnO = 0.222 mol ZnS

2 moles ZnO = 3 moles Oxygen

0.222 mol ZnO = 0.334 mol of Oxygen

Since, both the reactants are in enough concentration,

The theoretical yield can be calculated with any of the reactants. The theoretical yield of ZnO from 46.5 grams of ZnS can be given as:

1 mole ZnS = 1 mol ZnO

0.477 mol of ZnS = 0.477 mol of ZnO.

The mass of 0.477 mol of ZnO:

Mass = moles molecular weight

Mass of ZnO = 0.477 81.38

Mass of ZnO = 38.818 grams.

The theoretical yield of ZnO = 38.818 g.

The actualyield of ZnO = 18.14 g.

Percentage yield =

Percentage yield = 46.74 %

The percentage yield of the reaction has been 46.74%.

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