Question

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 6.19 g of sucrose and is the limiting reactant, what is the theoretical of H2O?

Answer 1

Answer on Question #47532, Chemistry, Other

Question: 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.

$$\mathrm{C}12\mathrm{H}22\mathrm{O}11(\mathrm{s}) + 12\mathrm{O}2(\mathrm{g}) \rightarrow 12\mathrm{CO}2(\mathrm{g}) + 11\mathrm{H}2\mathrm{O}(\mathrm{g})$$

Using the equation above, if the marshmallow contains $6.19\mathrm{g}$ of sucrose and is the limiting reactant, what is the theoretical of H2O?

Answer: From an equation of chemical reaction we see that from 1 mole of C12H22O11 we get 11 mole of H2O.

$$\begin{array}{l} n(\mathrm{C}12\mathrm{H}22\mathrm{O}11) = 6.19\mathrm{g}/342 = 0.02\mathrm{mole} \\ n(\mathrm{C}12\mathrm{H}22\mathrm{O}11)/n(\mathrm{H}2\mathrm{O}) = 1/11 = 0.02\mathrm{mole}/\mathrm{x}, \text{ so } \mathrm{x} = 11*0.02 = 0.22\mathrm{mole} \text{ of H2O} \\ V(\mathrm{H}20) = 0.22*22.4 = 4.93\mathrm{L} \\ \end{array}$$

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