posted by Sete Odhiambo .
Table salt, NaCl(s), and sugar, C12H22O11(s), are accidentally mixed. A 5.00-g sample is burned, and 2.50 g of CO2(g) is produced. What is the mass percentage of the table salt in the mixture
C12H22O11 + 12O2 ==> 12CO2 + 11H2O
Convert 2.50 g CO2 to moles grams sugar. The easy way is 2.50g CO2 x (molar mass sugar/12*molar mass CO2) = about 1.7 g (but you need to be more accurate).
Then 5.00g - 1.7g sugar = grams NaCl.
%NaCl = (mass NaCl/mass sample)*100 = ??
DrBob222's solution worked. The answer for 2.20g CO2 from 4g sample is 64.35%. Hope this helps someone with their Sapling HW.