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.

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To find the mass percentage of table salt in the mixture, you need to calculate the amount of table salt and the total mass of the mixture.

First, we need to find the amount of table salt (NaCl) by using stoichiometry. We know that 2.50 g of CO2 gas is produced, and CO2 is one of the products when sugar is burned. By balancing the equation for the combustion of sugar, we can determine the molar ratio between CO2 and sugar:

C12H22O11 + 12 O2 → 12 CO2 + 11 H2O

From the balanced equation, we can see that for every 12 moles of CO2 produced, we have 1 mole of sugar. We can convert the grams of CO2 to moles of CO2 using the molar mass of CO2:

2.50 g CO2 × (1 mol CO2 / 44.01 g CO2) = 0.0568 mol CO2

Since each mole of CO2 corresponds to one mole of sugar, we have 0.0568 mol of sugar.

Next, we need to calculate the mass of sugar:
0.0568 mol sugar × (342.3 g sugar / 1 mol sugar) = 19.43 g sugar

Now we can find the mass of table salt by subtracting the mass of sugar from the total mass of the mixture:
Total mass of mixture - Mass of sugar = Mass of table salt

Given that the total mass of the mixture is 5.00 g, we can substitute the values into the equation:
5.00 g - 19.43 g = -14.43 g

Since the mass of table salt cannot be negative, we can conclude that there was no table salt in the original mixture. Therefore, the mass percentage of table salt is 0%.