The combustion of hexyne, C6H10 is represented by the given equation:C6H10 + O2 --> CO2 + H2OIf 358 grams of C6H10 is available. If all C6H10 will be fully consumed, how many grams of CO2 will be produced for complete combustion assuming no excess O2 is supplied?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 358 grams of C6H10 is available. If all C6H10 will be fully consumed, how many grams of CO2 will be produced for complete combustion assuming no excess O2 is supplied?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 need to follow these steps:
- Calculate the molar mass of C6H10, CO2, and O2.
- Use stoichiometry to find the ratio of moles of C6H10 to CO2.
- Use the given mass of C6H10 to find the number of moles.
- Use the ratio from step 2 to find the number of moles of CO2.
- Convert the moles of CO2 to grams.
Step 1: Calculate the molar mass of C6H10, CO2, and O2.
- Molar mass of C6H10 = (612 g/mol) + (101 g/mol) = 72 g/mol + 10 g/mol = 82 g/mol
- Molar mass of CO2 = (112 g/mol) + (216 g/mol) = 12 g/mol + 32 g/mol = 44 g/mol
Step 2: Use stoichiometry to find the ratio of moles of C6H10 to CO2.
From the balanced chemical equation, we can see that 1 mole of C6H10 produces 6 moles of CO2.
Step 3: Use the given mass of C6H10 to find the number of moles.
Number of moles of C6H10 = mass / molar mass = 358 g / 82 g/mol = 4.37 moles
Step 4: Use the ratio from step 2 to find the number of moles of CO2.
Number of moles of CO2 = 4.37 moles of C6H10 * 6 moles of CO2/1 mole of C6H10 = 26.22 moles of CO2
Step 5: Convert the moles of CO2 to grams.
Mass of CO2 = number of moles * molar mass = 26.22 moles * 44 g/mol = 1153.68 g
So, 1153.68 grams of CO2 will be produced for complete combustion of 358 grams of C6H10.
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