A student reacts 20.0 grams of sodium hydroxide with hydrochloric acid and obtains 25.0 grams of sodium chloride. What is the percent yield?
You must first solve for the theoretical yield and you do that the same way you did for the KClO3 example I gave you in your post just above. Then
%yield = (actual yield/theoretical yield)*100 = ??
%CuS04.5H20
that Cu=63.5, S=32, 0(x9)=16*9=144 , H(*10)=1*10=10 that Cu+S+0+H=249,5 g
63.5/249.5*100%=25.5% what that is????
???
Actual Yield weight of copper after dry (g) ?????????
%cu= ????/
%Yield =Actual yield/theoretical Yield*100 ????????????/
That was experiment at school
To calculate the percent yield, we need to compare the actual yield (the amount of sodium chloride obtained) to the theoretical yield (the amount of sodium chloride that would be obtained if the reaction went to completion).
First, we need to determine the theoretical yield of sodium chloride based on the balanced chemical equation for the reaction:
2NaOH + HCl -> NaCl + H2O
Looking at the coefficients in the balanced equation, we see that 2 moles of sodium hydroxide react to produce 1 mole of sodium chloride. So, we need to convert the grams of sodium hydroxide to moles, and then calculate the moles of sodium chloride that should have been obtained.
Molar mass of NaOH:
Sodium (Na) = 22.99 g/mol
Oxygen (O) = 16.00 g/mol
Hydrogen (H) = 1.01 g/mol
Total molar mass of NaOH = 22.99 + 16.00 + 1.01 = 40.00 g/mol
Converting grams of NaOH to moles:
Moles of NaOH = Mass of NaOH / Molar mass of NaOH
Moles of NaOH = 20.0 g / 40.00 g/mol = 0.50 mol
Since the ratio between sodium hydroxide and sodium chloride is 2:1, the moles of sodium chloride should be half of the moles of sodium hydroxide.
Moles of NaCl (theoretical yield) = 0.50 mol / 2 = 0.25 mol
Now, we can calculate the percent yield using the formula:
Percent yield = (Actual yield / Theoretical yield) x 100
Actual yield = 25.0 g (given in the question)
Percent yield = (25.0 g / 0.25 mol) x 100
= 1000 %
Therefore, the percent yield is 1000%.