How many grams of octane (C8H18) must be burned to produce 300.0g of CO2?
1 answer:
Answer:
Explanation:
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In this case, according to the given combustion reaction of octane, it is possible for us to perform the stoichiometric method in order to calculate the mass of octane that is required to consume 300.0 g of oxygen by considering the 2:25 mole ratio, and the molar masses of 114.22 g/mol and 32.00 g/mol respectively:
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Answer:
Similarities: both state the mass of chemical species and they have the same numerical value
Differences: molecular mass refers to one single molecule and molar mass refers to one mole of a molecule
Explanation:
The molecular mass is the value of the mass of each molecule and it is measured in mass units (u). It is calculated adding the mass of each atom of the molecule.
The molar mass is the value of the mass of one mole of molecules, which means the mass of 6.022140857 × 10²³ molecules. The unit is g/mol.
For example, we can consider the methane molecule, which has the chemical formula of CH₄:
Molecular mass CH₄ = C mass + 4 x (H mass)
Molecular mass CH₄ = 12.01 + 4 x (1.01)
Molecular mass CH₄ = 16.05 u
Now to calculate the molar mass we multiply the value of the molecular mass by the Avogadro number and convert the units to g/mol:
Molar mass CH₄: 16.05 x x 6.022140857 × 10²³ mol⁻¹
Molecular mass CH₄ = 16.05 g / mol