A TECHNICIAN PLACES 3.0 mol OF SULFUR TRIOXIDE IN A GAS SEALED 500 mL CONTAINER AND ALLOWED THE REACTION TO ESTABLISH THE FOLLOWING EQUALIBRIUM
2SO3(g) ---> 2SO2(g) + O2(g)
P.S THE ARROW IS SUPPOSED TO BE A REVERSE AND FORWARD REACTION SYMBOL.
AT EQUALIBRIUM THE SYSTEM WAS DETERMINED TO CONTAIN 0.95 MOL OF OXYGEN GAS.
WHAT IS THE PERCENT YIELD?
5 years ago
We can answer this question by establishing a proportion between the number of moles of sulfur trioxide and oxygen gas given in the reaction:
2 moles of sulfur trioxide produces 1 mol of oxygen gas, theoretically.
Hence, how many moles of oxygen gas can we produce if we place 3 moles of sulfur trioxide?
$2moles\left(SO_3\right)=1mol\left(O_2\right)$2moles(SO3)=1mol(O2)
$3moles\left(SO_3\right)=x$3moles(SO3)=x
If we cross multiply the above proportion:
$\left(3\right)\left(1\right)=\left(2\right)x$(3)(1)=(2)x
$x=\frac{3}{2}=1.5mol\left(O_2\right)$x=32 =1.5mol(O2)
Hence, we can theoretically produce 1.5 mol of oxygen gas. The exercise told us that we actually produce 0.95 mol of oxygen gas. If we divide this amount by the theoretical amount (1.5 mol), we are going to obtain the yield of the reaction:
$Yield=\frac{0.95}{1.5}=0.63=63\%$Yield=0.951.5 =0.63=63%
The percent yield of this reaction is approximately 63%.
5 years ago
Answered By Leonardo F
We can answer this question by establishing a proportion between the number of moles of sulfur trioxide and oxygen gas given in the reaction:
2 moles of sulfur trioxide produces 1 mol of oxygen gas, theoretically.
Hence, how many moles of oxygen gas can we produce if we place 3 moles of sulfur trioxide?
$2moles\left(SO_3\right)=1mol\left(O_2\right)$2moles(SO3)=1mol(O2)
$3moles\left(SO_3\right)=x$3moles(SO3)=x
If we cross multiply the above proportion:
$\left(3\right)\left(1\right)=\left(2\right)x$(3)(1)=(2)x
$x=\frac{3}{2}=1.5mol\left(O_2\right)$x=32 =1.5mol(O2)
Hence, we can theoretically produce 1.5 mol of oxygen gas. The exercise told us that we actually produce 0.95 mol of oxygen gas. If we divide this amount by the theoretical amount (1.5 mol), we are going to obtain the yield of the reaction:
$Yield=\frac{0.95}{1.5}=0.63=63\%$Yield=0.951.5 =0.63=63%
The percent yield of this reaction is approximately 63%.