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Liquid octane CH3CH26CH3 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O. Suppose 2.3 g of octane is mixed with 12.4 g of oxygen. Calculate the minimum mass of octane that could be left over by the chemical reaction. Round your answer to 2 significant digits.

User Glindste
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Answer: Octane will be used completely.

Step-by-step explanation:

To calculate the moles :


\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}


\text{Moles of octane}=(2.3g)/(114g/mol)=0.0202moles


\text{Moles of oxygen}=(12.4g)/(32g/mol)=0.388moles

The balanced chemical reaction will be


2C_8H_(18)+25O_2(g)\rightarrow 16CO_2(g)+18H_2O(g)

According to stoichiometry :

2 moles of octane require = 25 moles of
O_2

Thus 0.0202 moles of octane will require=
(25)/(2)* 0.0202=0.2525moles of
O_2

Thus octane is the limiting reagent as it limits the formation of product and
O_2 is the excess reagent.

Thus octane will be used completely.

User TheCyberXP
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