The reaction: 2C2H2+5O2->5O2->4CO2+2H2O is performed by mixing 175 grams of C2H2 and 175 grams of O2 to produce 127 grams of CO2
1)Limiting reagent
2)Theretical yeield of CO2
3)Percent yield of CO2
To determine the limiting reagent, theoretical yield of CO2, and percent yield of CO2, we need to follow a step-by-step approach.
1) Limiting reagent:
To find the limiting reagent, we compare the number of moles of each reactant to the stoichiometry of the balanced equation. The reactant that produces fewer moles of CO2 is the limiting reagent.
Let's calculate the number of moles of C2H2 and O2:
- C2H2: The molar mass of C2H2 is 26.04 g/mol.
Moles of C2H2 = mass of C2H2 / molar mass of C2H2 = 175 g / 26.04 g/mol
- O2: The molar mass of O2 is 32.00 g/mol.
Moles of O2 = mass of O2 / molar mass of O2 = 175 g / 32.00 g/mol
Now, to compare the mole ratios with the balanced chemical equation:
- The reaction is: 2C2H2 + 5O2 -> 4CO2 + 2H2O
- The stoichiometric ratio of C2H2 to CO2 is 2:4, or simplified, 1:2.
- The stoichiometric ratio of O2 to CO2 is 5:4, or simplified, 5:4.
Now, multiply the moles of each reactant by their stoichiometric ratios:
- Moles of C2H2 x stoichiometric ratio with CO2 = Moles of C2H2 x 2/1 (since the ratio is 1:2)
- Moles of O2 x stoichiometric ratio with CO2 = Moles of O2 x 5/4 (since the ratio is 5:4)
Compare the products of the calculations:
- If the moles of C2H2 x 2/1 are lower, C2H2 is the limiting reagent.
- If the moles of O2 x 5/4 are lower, O2 is the limiting reagent.
In this case, calculate both and compare.
2) Theoretical yield of CO2:
To calculate the theoretical yield of CO2, we need to determine the number of moles of CO2 produced from the limiting reagent and then convert it to grams using the molar mass of CO2.
- Using the balanced equation, the stoichiometric ratio between CO2 and C2H2 is 4:2, or simplified, 2:1.
- Calculate the moles of CO2 produced from the moles of the limiting reactant using this ratio.
- Moles of CO2 = Moles of limiting reagent x stoichiometric ratio with CO2 = Moles of limiting reagent x 2/1
Then, to calculate the theoretical yield of CO2:
- Theoretical yield of CO2 = Moles of CO2 x molar mass of CO2
3) Percent yield of CO2:
The percent yield is calculated by dividing the actual yield of CO2 (given as 127 grams) by the theoretical yield of CO2, then multiplying by 100.
- Percent yield of CO2 = (Actual yield of CO2 / Theoretical yield of CO2) x 100
Now, using the steps explained, you can find the answers to the limiting reagent, theoretical yield of CO2, and percent yield of CO2 for the given reaction using the given data.
To find the limiting reagent, we need to determine which reactant will be completely consumed first.
1) Limiting reagent:
To do this, we can compare the moles of C2H2 and O2:
- The molar mass of C2H2 (acetylene) is 26.04 g/mol.
- The molar mass of O2 (oxygen) is 32.00 g/mol.
First, we need to convert the given masses of C2H2 and O2 to moles:
- Moles of C2H2 = (175 g) / (26.04 g/mol)
- Moles of O2 = (175 g) / (32.00 g/mol)
Next, we look at the balanced equation:
2 C2H2 + 5 O2 -> 4 CO2 + 2 H2O
From the equation, we can see that 2 moles of C2H2 react with 5 moles of O2 to produce 4 moles of CO2.
Now, let's calculate the moles of C2H2 and O2 needed to react completely:
- Moles of C2H2 needed = 2/5 * (moles of O2)
If the moles of C2H2 needed is less than the moles of C2H2 given, then C2H2 is the limiting reagent. Otherwise, O2 is the limiting reagent.
2) Theoretical yield of CO2:
To find the theoretical yield of CO2, we need to calculate the moles of CO2 that can be produced from the limiting reagent.
The molar mass of CO2 (carbon dioxide) is 44.01 g/mol.
Using the balanced equation, we know that 2 moles of C2H2 react to produce 4 moles of CO2.
The moles of CO2 produced = (moles of C2H2) * (4 moles CO2 / 2 moles C2H2)
Now, we can calculate the theoretical yield of CO2:
Theoretical yield of CO2 = (moles of CO2 produced) * (molar mass of CO2)
3) Percent yield of CO2:
The percent yield is the actual yield of a reaction divided by the theoretical yield, multiplied by 100%.
The actual yield is given as 127 grams.
Percent yield of CO2 = (actual yield / theoretical yield) * 100%
change the mass to moles for all three.
Now divide the moles of C2H2, O2, and CO2 by the coefficents 2,5, 4
Which is the lowest number? That is the limiting reageant. WHY?
For yield, how many moles of the limiting reageant so you have. You should get the ratio of coefficents of CO2, and the ratio of the limitng reagent. Convert that to grams.
3. Percent yield. divide the actual yield by the theoritical.