Asked by Chelsea
Table salt NaCl and sugar C12H22O11 are accidentally mixed. A 6.00 g sample is burned and 2.90 g of CO2 is produced. What was the mass percentage of the table salt mixture?
Could you please show me all the way through if you can?
Could you please show me all the way through if you can?
Answers
Answered by
DrBob222
NaCl doesn't burn. Sugar does.
C12H22O11 + 12O2 ==> 12CO2 + 11H2O
Convert g CO2 to mols sugar.
2.90g CO2 x (molar mass sugar/12*molar mass CO2) = 2.90 x (342/12*44) = about 1.9
NaCl + sugar = 6.00
mass sugar = -1.9
mass NaCl = 6-1.9 about 4.1
%NaCl = (mass NaCl/mass sample)*100 = ?
%NaCl = about (4.1/6)*100 = about 68.7%
You should go through and do it more accurately. I've just estimated here and ther.
C12H22O11 + 12O2 ==> 12CO2 + 11H2O
Convert g CO2 to mols sugar.
2.90g CO2 x (molar mass sugar/12*molar mass CO2) = 2.90 x (342/12*44) = about 1.9
NaCl + sugar = 6.00
mass sugar = -1.9
mass NaCl = 6-1.9 about 4.1
%NaCl = (mass NaCl/mass sample)*100 = ?
%NaCl = about (4.1/6)*100 = about 68.7%
You should go through and do it more accurately. I've just estimated here and ther.