When 28.0 g of acetylene reacts with hydrogen, 24.5 g of ethane is produced. What is the percent yield of C2H6 for the reaction?
1 answer:
Answer:
Explanation:
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In this case, since no information about the reacting hydrogen is given, we can assume that it completely react with the 28.0 g of acetylene to yield ethane. In such a way, via the 1:1 mole ratio between acetylene (molar mass = 26 g/mol) and ethane (molar mass = 30 g/mol), we compute the yielded grams, or the theoretical yield of ethane as shown below:
Hence, by knowing that the percent yield is computed via the actual yield (24.5 g) over the theoretical yield, we obtain:
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Here we have to get the of the reaction at 520 K temperature.
The of the reaction is 1.705 atm
We know the relation between and
is
, where
= The equilibrium constant of the reaction in terms of partial pressure,
= The equilibrium constant of the reaction in terms of concentration and N = number of moles of gaseous products - Number of moles of gaseous reactants.
Now in this reaction, PCl₃ + Cl₂ ⇄ PCl₅
Thus number of moles of gaseous product is 1, and number of moles of gaseous reactants are 2. Thus N = |1 - 2| = 1 mole
The given value of is 4.0×10⁻²
The molar gas constant, R = 0.082 L. Atm. mol⁻¹. K⁻¹ and temperature, T = 520 K.
On plugging the values in the equation we get,
Or, = 1.705 atm
Thus, the of the reaction is 1.705 atm