Now you've learned everything! The goal of this site is to acquaint avid chemistry lovers with all the tools needed to express their love of practicing chemistry. With all the many methods you've learned its time to tackle the ultimate type of problem. First we will run through how to combine each technique you've learned and then I will send you off into the world of chemicals with one last test.
Lets try it together first
For the balanced equation 2Na + 2H2O --> 2NaOH + H2 , if 42 grams of Na is reacted with 59 grams of H2O, and produce a 77% yield of H2, how many grams of H2 would be produced?
Alright. First lets find the limiting reagent
42g/23 (GFW of Na) x 2=3.65 mol versus 59g/18 (GFW of H2O) x 2=6.555 mol
Na is smaller so it is the limiting reagent
Now the theoretical yield
The ratio is 2:1 or 1:1/2
1/2 x 42g/23 (moles of Na= .913 mol
Theoretical yield=.913 mol x 2 (GFW of H2)=1.826g
Now we find the grams produced by using the percent yield
x grams (actual yield)/1.826g (theoretical yield) x 100=77
.77g x 1.826g=x
x= 1.40602 g
1.40602 grams of H2 were actually produced
Ready for your send off.
Lets try it together first
For the balanced equation 2Na + 2H2O --> 2NaOH + H2 , if 42 grams of Na is reacted with 59 grams of H2O, and produce a 77% yield of H2, how many grams of H2 would be produced?
Alright. First lets find the limiting reagent
42g/23 (GFW of Na) x 2=3.65 mol versus 59g/18 (GFW of H2O) x 2=6.555 mol
Na is smaller so it is the limiting reagent
Now the theoretical yield
The ratio is 2:1 or 1:1/2
1/2 x 42g/23 (moles of Na= .913 mol
Theoretical yield=.913 mol x 2 (GFW of H2)=1.826g
Now we find the grams produced by using the percent yield
x grams (actual yield)/1.826g (theoretical yield) x 100=77
.77g x 1.826g=x
x= 1.40602 g
1.40602 grams of H2 were actually produced
Ready for your send off.