The combustion of hexyne, C6H10 is represented by the given equation:C6H10 + O2 --> CO2 + H2OIf 338 grams of C6H10 is available. If all C6H10 will be fully consumed, how many grams of O2 is needed for complete combustion?Write your final answer in two decimal places.Use the following mass numbers:C - 12 g/molH - 1 g/molO - 16 g/mol
Question
The combustion of hexyne, C6H10 is represented by the given equation:C6H10 + O2 --> CO2 + H2OIf 338 grams of C6H10 is available. If all C6H10 will be fully consumed, how many grams of O2 is needed for complete combustion?Write your final answer in two decimal places.Use the following mass numbers:C - 12 g/molH - 1 g/molO - 16 g/mol
Solution
To solve this problem, we first need to balance the chemical equation. The balanced equation for the combustion of hexyne is:
C6H10 + 9O2 --> 6CO2 + 5H2O
Next, we calculate the molar mass of C6H10 and O2.
The molar mass of C6H10 is (612 g/mol) + (101 g/mol) = 82 g/mol. The molar mass of O2 is 2*16 g/mol = 32 g/mol.
According to the balanced equation, 1 mole of C6H10 reacts with 9 moles of O2.
So, if we have 338 g of C6H10, the number of moles is 338 g / 82 g/mol = 4.12 moles.
Since 1 mole of C6H10 reacts with 9 moles of O2, 4.12 moles of C6H10 will react with 4.12 * 9 = 37.08 moles of O2.
Finally, we convert the moles of O2 to grams using the molar mass.
The mass of O2 needed is 37.08 moles * 32 g/mol = 1186.56 g.
So, 1186.56 grams of O2 are needed for the complete combustion of 338 grams of C6H10.
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