Question 6.8: The human body converts the glucose, C6H12O6 , contained in ......
The human body converts the glucose, C_{6}H_{12}O_{6}, contained in foods to carbon dioxide, CO_{2}, and water, H_{2}O. The chemical equation for the chemical reaction is
\quad\quad\quad\quad\quad\quad C_{6}H_{12}O_{6} + {6}O_{2} → {6}CO_{2} + {6}H_{2}O
Assume a person eats a candy bar containing 14.2 g (1/2 oz) of glucose. How many grams of water will the body produce from the ingested glucose, assuming all of the glucose undergoes reaction?
Step 1: The given quantity is 14.2 g of glucose. The desired quantity is grams of water.
\quad\quad\quad\quad 14.2 g C_{6}H_{12}O_{6} = ? g H_{2}O
In terms of Figure 6.9, this is a “grams of A” to “grams of B” problem.
Step 2: Using Figure 6.9 as a road map, we determine that the pathway for this problem is
The mathematical setup for this problem is
14.2 \cancel{g C_{6}H_{12}O_{6}}\times (\frac{1 \cancel{mole C_{6}H_{12}O_{6}}}{180.18 \cancel{g C_{6}H_{12}O_{6}}})\times (\frac{6 \cancel{moles H_{2}O}}{1 \cancel{mole C_{6}H_{12}O_{6}}})\times (\frac{18.02 \cancel{g H_{2}O}}{1 \cancel{mole H_{2}O}})\\ \quad\quad g C_{6}H_{12}O_{6}\longrightarrow moles C_{6}H_{12}O_{6}\longrightarrow moles H_{2}O\longrightarrow g H_{2}OThe 180.18 g in the first conversion factor is the molar mass of glucose, the 6 and 1 in the second conversion factor are the coefficients, respectively, of H_{2}O and C_{6}H_{12}O_{6} in the balanced chemical equation, and the 18.02 g in the third conversion factor is the molar mass of H_{2}O.
Step 3: The solution to the problem, obtained by doing the arithmetic after all the numerical factors have been collected, is
