QuestionQuestion

When left over a flame a marshmallow is consumed in a combustion reaction. Assuming a marshmallow is made of entirely sucrose, C12H22O11, and after combustion no residual solid remained, the following chemical equation describes this reaction.

C12H22O11(s) + 12 O2(g) → 12 CO2(g) + 11 H2O(g)

Using the equation above, if the marshmallow contains 6.8 g of sucrose and is the limiting reactant, what is the theoretical yield, in g, of H2O?

1. Using the chemical equation in Question 1 above, if a second marshmallow was measured and contained 6.94 g of sucrose and is the limiting reactant, what is the theoretical yield, in g, of CO2?
2. Using the chemical equation in Question 1 above, if 5.91 g of C12H22O11 reacts in excess oxygen gas and the actual yield of H2O was 2.708 g, calculate the percent yield of the reaction. Hint: You'll need to calculate the theoretical yield, in g, of H2O first.
3. The percent composition of an unknown compound was 69.5 % Fe and 30.5 % O. What is the empirical formula for this compound?
4. The percent composition of an unknown compound was 40.00 % C, 6.666 % H, and 53.33 % O. The unknown compound also has a molecular weight of 60.06 g/mol. What is the molecular formula for this compound?

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