Show below is the combustion of benzene, C6H6.C6H6 + O2--> CO2 + H2O1,001 grams of benzene are supplied and some excess air. If3,077 grams of carbon dioxide are produced, compute for the percentage yield.Write your final answer in two decimal places.Use the following mass numbers:C - 12 g/molH - 1 g/molO - 16 g/mol
Question
Show below is the combustion of benzene, C6H6.C6H6 + O2--> CO2 + H2O1,001 grams of benzene are supplied and some excess air. If3,077 grams of carbon dioxide are produced, compute for the percentage yield.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 calculate the percentage yield, we first need to determine the theoretical yield of CO2 from the combustion of benzene.
Step 1: Calculate the molar mass of benzene (C6H6) The molar mass of benzene is (612 g/mol for C) + (61 g/mol for H) = 78 g/mol.
Step 2: Calculate the moles of benzene used The moles of benzene used is the mass of benzene divided by its molar mass. So, 1.001 g / 78 g/mol = 0.01283 mol.
Step 3: Determine the stoichiometry of the reaction From the balanced chemical equation, we can see that 1 mol of benzene (C6H6) produces 6 mol of CO2.
Step 4: Calculate the theoretical yield of CO2 The theoretical yield of CO2 is therefore 0.01283 mol * 6 = 0.077 mol.
Step 5: Convert this to grams The molar mass of CO2 is (12 g/mol for C) + (2*16 g/mol for O) = 44 g/mol. So, the theoretical yield in grams is 0.077 mol * 44 g/mol = 3.388 g.
Step 6: Calculate the percentage yield The percentage yield is the actual yield (the mass of CO2 produced) divided by the theoretical yield, multiplied by 100%. So, (3.077 g / 3.388 g) * 100% = 90.81%.
Therefore, the percentage yield of the reaction is 90.81%.
Similar Questions
Show below is the combustion of hexane, C6H14.C6H14 + O2--> CO2 + H2OIf 444 grams of hexane and 2,038 grams of air are supplied, compute for the percentage excess.Write your final answer in two decimal places.Use the following mass numbers:C - 12 g/molH - 1 g/molO - 16 g/mol
Given:C2H6(g) + O2(g) ⟶ CO2(g) + H2O(g)If 319 grams of ethane and 1,597 grams of air are supplied, compute for the % excess.Write your final answer in two decimal places.Use the following mass numbers:O - 16C - 12H - 1
A mass of 8.15 g C2H4(g) reacts with excess oxygen. If 16.2 g CO2(g) is collected, what is the percent yield of the reaction? C2H4(g) + 3O2(g) → 2CO2(g) + 2H2O(g)Group of answer choices41.0%31.7%57.1%63.3%
The combustion of hexyne, C6H10 is represented by the given equation:C6H10 + O2 --> CO2 + H2OIf 403 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
If 4.750 g of CO2(g) was obtained from the combustion of 1.638 g of propane, what was the percent yield?
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.