As we exercise, our bodies metabolize glucose, converting it to C O 2 and H 2 O

Question

As we exercise, our bodies metabolize glucose, converting it to C O 2 and H 2 O , to supply the energy necessary for physical activity. The simplified reaction is: C 6 H 12 O 6 (aq)+6 O 2 (g) 6C O 2 (g)+6 H 2 O(l)+678 kcal (2840 kJ).

How many grams of water would have to be evaporated as sweat to remove the heat generated by the metabolism of 3 mol of glucose? (See Chemistry in Action: Regulation of Body Temperature on p. 195 in the textbook).

Express your answer to four significant figures and include the appropriate units.

Explanation / Answer

C6H12O6 (aq) + 6O2 (g) 6CO2 (g) + 6 H2O(l) + 2840 kJ

1 mole of glucose produces 6 moles of water

3x1 moles of glucose produces 3x6=18 moles of water

Molar mass of water,H2O = ( 2x1) + 16 = 18 g/mol

So mass of water evaporated is , m = number of moles of water x molar mass

                                                    = 18 mol x 18 g/mol

                                                    = 324.0 g

Therefore the mass of evaporated is 324.0 g

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