If 10.5 g of hydrogen, H2, were mixed with 6.51 g of acetylene, C2H2, and allowed to react according to the following equation, what is the theoretical yield of ethane,C2H6 produced? If 7.00g of ethane, C2H6 were recovered at the end of the reaction , what is the percent yield.
C2H2(g)+2H2(g)=C2H6(g)
This is a limiting reagent problem.
mols H2 = grams/molar mass
mols C2H2 = grams/molar mass.
Convert mols H2 to mols C2H6 using the coefficients in the balanced equation.
Do the same for mols C2H2.
It is likely that the two values will not agree which means one of them is wrong. The correct answer in limiting reagent problems is ALWAYS the smaller value and the reagent producing that value is the limiting reagent.
Using the smaller value convert to grams. g = mols x molar mass. This is the theoretical yield.
% yield = (actual yi8eld/theor yield)*100 = ?
To calculate the theoretical yield of ethane, C2H6, we need to determine the limiting reactant in the reaction. The limiting reactant is the reactant that is completely consumed and determines the maximum amount of product that can be formed.
To find the limiting reactant, we compare the moles of each reactant with their respective coefficients in the balanced equation.
First, we need to convert the given masses of hydrogen (H2) and acetylene (C2H2) to moles using their molar masses.
The molar mass of hydrogen (H2) is 2.02 g/mol, and the molar mass of acetylene (C2H2) is 26.04 g/mol.
For hydrogen (H2):
Moles of H2 = Mass of H2 / Molar mass of H2
Moles of H2 = 10.5 g / 2.02 g/mol = 5.20 mol
For acetylene (C2H2):
Moles of C2H2 = Mass of C2H2 / Molar mass of C2H2
Moles of C2H2 = 6.51 g / 26.04 g/mol = 0.250 mol
Now, let's compare the moles of each reactant with their coefficients in the balanced equation:
C2H2: 0.250 mol --> coefficient: 1
H2: 5.20 mol --> coefficient: 2
From the balanced equation, we can see that the ratio of C2H2 to H2 is 1:2. This means that for every 1 mole of C2H2, 2 moles of H2 are required.
Since the moles of H2 (5.20 mol) are greater than twice the moles of C2H2 (0.250 mol * 2 = 0.500 mol), H2 is in excess, and C2H2 is the limiting reactant.
Now, we can calculate the moles of ethane (C2H6) produced from the limiting reactant:
From the balanced equation: 1 mole of C2H2 yields 1 mole of C2H6
Moles of C2H6 = Moles of C2H2 (limiting reactant) = 0.250 mol
Finally, we can calculate the theoretical yield of ethane (C2H6) in grams:
Mass of C2H6 = Moles of C2H6 * Molar mass of C2H6
Mass of C2H6 = 0.250 mol * 30.07 g/mol = 7.52 g
Therefore, the theoretical yield of ethane (C2H6) is 7.52 g.
To calculate the percent yield, we need to compare the actual yield (7.00 g) to the theoretical yield (7.52 g) and calculate the percentage:
Percent Yield = (Actual Yield / Theoretical Yield) * 100
Percent Yield = (7.00 g / 7.52 g) * 100 = 93.1%
Therefore, the percent yield of ethane (C2H6) is 93.1%.