FIRST MY WAY
Na = 23 grams / mol
Cl = 35.5 grams/mol
so
NaCl is 23 + 35.5 g/mol = 58.5 g/mol
so in that mol of Na Cl which weighs 58.5 g
you have one mol of Cl which weighs 35.5 g
so
what is 100 * 35.5 / 58.5 = 60.7 %
now what they did (silly)
100 * 7.17/10.3 = 69.6 %
3.13 g of sodium (Na) react with 7.17 g of chlorine (Cl2) to produce 10.3 g of sodium chloride (NaCl).
What is the mass percent of chlorine in sodium chloride?
50.0% is chlorine.
30.3% is chlorine.
69.6% is chlorine.
43.7% is chlorine.
I do not believe it is 30.3 or 43.7, I believe it is 69.6
2 answers
Not only is it silly; it is not possible for those numbers.
Starting with 3.13 g Na you should, at 100% yield, obtain *2*58.5/2*23 = 7.96 g NaCl but they said 10.3. Can't be, that's better than 100% yield. OR,
starting with 7.17 g Cl2 you would get, at 100% yield, 7.17 x (2*58.5/71) = 11.8 g. Can't be unless the reaction was < 100%. Dead cat on the line somewhere. 60.7% chlorine in NaCl is correct.(71/2*58.5 )100 = 60.68% which rounds to 60.7%. OR,
The limiting reagent is Na @ 3.13 g. How much Cl2 is needed to react @ 100%. That is 3.13*71/2*23 = 4.83.(obviously there is excess Cl2)
Total NaCl is g Na + g Cl2 = 3.13 + 4.83 = 7.96. Now the coup de grace is
%Chlorine = (g chlorine/g NaCl total)100 = 100 x 4.83/7.95 = 60.68 or 60.7%. Voila!
Starting with 3.13 g Na you should, at 100% yield, obtain *2*58.5/2*23 = 7.96 g NaCl but they said 10.3. Can't be, that's better than 100% yield. OR,
starting with 7.17 g Cl2 you would get, at 100% yield, 7.17 x (2*58.5/71) = 11.8 g. Can't be unless the reaction was < 100%. Dead cat on the line somewhere. 60.7% chlorine in NaCl is correct.(71/2*58.5 )100 = 60.68% which rounds to 60.7%. OR,
The limiting reagent is Na @ 3.13 g. How much Cl2 is needed to react @ 100%. That is 3.13*71/2*23 = 4.83.(obviously there is excess Cl2)
Total NaCl is g Na + g Cl2 = 3.13 + 4.83 = 7.96. Now the coup de grace is
%Chlorine = (g chlorine/g NaCl total)100 = 100 x 4.83/7.95 = 60.68 or 60.7%. Voila!
2 answers