If 2CH2Cl2+3O2=>2CO2+2H2O+2Cl2, if 13.0 grams of CH2Cl2 were reacted with 48.6 grams of O2, first find the limiting reagent, then the theoretical yield of Cl2 and knowing that 32 is the actual yield, find the percentage yield.
First find the GFW of the reactants in question.
CH2Cl2: 84
O2: 32
Next, we divide the number of grams of each reactant by the GFW.
CH2Cl2: 13/84= 0.578
O2: 48.6/32= 1.51
We then multiply each result by the coefficient of the other reactant.
0.578 x 3= 1.734
11.51 x 2= 3.02
The limiting reagent is CH2Cl2. We take the coefficient of this and put it into a ratio with the coefficient of the desired product (Cl2)
2:2
Next, we divide this by the coefficient of the limiting reagent.
2/2= 1 : 2/2=1
Then we multiply the number on the right by the number of moles in the limiting reagent.
1 x 0.578= 0.578
We then multiply this by the GFW of the product.
0.578 x 70= 40.46
Next we plug this into the equation (Actual Yield)(Theoretical yield)x100
(32/40.46) x 100= 79.09%
And you're done!
First find the GFW of the reactants in question.
CH2Cl2: 84
O2: 32
Next, we divide the number of grams of each reactant by the GFW.
CH2Cl2: 13/84= 0.578
O2: 48.6/32= 1.51
We then multiply each result by the coefficient of the other reactant.
0.578 x 3= 1.734
11.51 x 2= 3.02
The limiting reagent is CH2Cl2. We take the coefficient of this and put it into a ratio with the coefficient of the desired product (Cl2)
2:2
Next, we divide this by the coefficient of the limiting reagent.
2/2= 1 : 2/2=1
Then we multiply the number on the right by the number of moles in the limiting reagent.
1 x 0.578= 0.578
We then multiply this by the GFW of the product.
0.578 x 70= 40.46
Next we plug this into the equation (Actual Yield)(Theoretical yield)x100
(32/40.46) x 100= 79.09%
And you're done!