(a) Balance this unbalanced equation by inspection: $$ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}+\mathrm{O}_{2} \longrightarrow \mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O} $$ Glucose (b) Express this reaction in words, using the word mole(s) wherever appropriate. (c) To produce 6 moles of \(\mathrm{H}_{2} \mathrm{O}\) from this reaction, how many grams of glucose and \(\mathrm{O}_{2}\) must you combine? (d) What is the theoretical yield in grams of \(\mathrm{CO}_{2} ?\) (e) Suppose you recover \(196.0 \mathrm{~g}\) of \(\mathrm{CO}_{2}\). What is the percent yield?

In order to balance the given unbalanced equation, we will need to adjust the coefficients of each molecule, so the number of atoms of each element on the reactant side equals the number of atoms of the same element on the product side. Given unbalanced equation: \[ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}+\mathrm{O}_{2} \longrightarrow \mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O} \] Balanced equation: \[ \mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6 + 6\mathrm{O}_2 \longrightarrow 6\mathrm{CO}_2 + 6\mathrm{H}_2\mathrm{O} \]

Using the balanced equation, we can express the reaction in words by identifying the reactants and products and mentioning their respective moles: "One mole of glucose reacts with six moles of oxygen to produce six moles of carbon dioxide and six moles of water."

Starting with the stoichiometric coefficients from the balanced equation, we can find out how many grams of glucose and O2 are needed to produce 6 moles of H2O: For 6 moles of H2O, the stoichiometry of the balanced equation shows that we need: 1 mole of glucose, from the balanced equation: Molar mass of glucose: \(C_6H_{12}O_6 = 6 × 12.01 + 12 × 1.01 + 6 × 16.00 = 180.18\mathrm{g/mol}\) Grams of glucose: \(1\mathrm{~mol} × 180.18\mathrm{~g/mol} = 180.18\mathrm{~g}\) 6 moles of O2, from the balanced equation: Molar mass of O2: \(O_2 = 2 × 16.00 = 32.00\mathrm{g/mol}\) Grams of O2: \(6\mathrm{~mol} × 32.00\mathrm{~g/mol} = 192.00\mathrm{~g}\) So, we need 180.18 g of glucose and 192.00 g of oxygen to produce 6 moles of water.

The theoretical yield of CO2 can be determined using the stoichiometry of the balanced equation and the molar mass of CO2: Molar mass of CO2: \(CO_2 = 12.01 + 2 × 16.00 = 44.01\mathrm{g/mol}\) 6 moles of CO2, from the balanced equation (since there's a 1:1 relationship between glucose and CO2): \(6\mathrm{~mol} × 44.01\mathrm{~g/mol} = 264.06\mathrm{~g}\) The theoretical yield of CO2 is 264.06 g.

To find the percent yield, we need to compare the actual yield (the amount of CO2 recovered) to the theoretical yield: Percent Yield = \(100 × \frac{\text{Actual Yield}}{\text{Theoretical Yield}}\) Given: Actual Yield = 196.0 g Percent Yield = \(100 × \frac{196.0\mathrm{~g}}{264.06\mathrm{~g}} ≈ 74.24\%\) The percent yield of the reaction is approximately 74.24%.