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 7.00 g of ethane, C2H6, were recovered at the end of the reaction, what is the percent yield?. C2H2(g)+2H2(g)=>C2H6(g)
Molar mass of acetylene is 35.01 g/mol
n=6.51/35.01= 0.1859 mol of C2H2
1 mol C2H2 0.1859 mol C2H2
__________= ________________
1 mol C2H6 X mol C2H6
x= 0.1859 mol C2H6
Molar mass of ethanol is 38.06 g/mol
m=(0.1859 mol)(38.06)= 7.07 g this is theoretical yield
% yield 7.00 g/7.07g x100%= 99% this is percentage yield
To find the theoretical yield of ethane (C2H6) produced, we need to use the stoichiometry of the balanced chemical equation. The stoichiometry tells us the mole ratio of reactants and products involved in the reaction.
1. Calculate the number of moles of hydrogen (H2):
- Mass of hydrogen (H2) = 10.5 g
- Molar mass of H2 = 2 g/mol (1 mole of H2 weighs 2 grams)
- Moles of H2 = Mass of H2 / Molar mass of H2
2. Calculate the number of moles of acetylene (C2H2):
- Mass of acetylene (C2H2) = 6.51 g
- Molar mass of C2H2 = 26.04 g/mol (2 carbons = 12.01 g/mol each, 2 hydrogens = 1.01 g/mol each)
- Moles of C2H2 = Mass of C2H2 / Molar mass of C2H2
3. Determine the limiting reactant:
- Compare the number of moles of hydrogen (H2) and acetylene (C2H2) to see which one is present in a lesser amount.
- The reactant that produces a lesser amount of product is the limiting reactant.
4. Calculate the theoretical yield of ethane (C2H6):
- Determine the mole ratio between acetylene (C2H2) and ethane (C2H6) from the balanced equation (1:1).
- Theoretical yield of C2H6 = Moles of limiting reactant (either H2 or C2H2) * Conversion factor from balanced equation.
5. Calculate the percent yield:
- Percent yield = (Actual yield / Theoretical yield) * 100
- Mass of ethane (C2H6) recovered = 7.00 g
These step-by-step calculations will help you find the theoretical yield of ethane (C2H6) produced and the percent yield.