i don\'t know if my other answers are right or not. Gen ChemIxam 3 A student wel

Question

i don't know if my other answers are right or not.

Gen ChemIxam 3 A student welghs 36.0 g solid glucose (MM180. g/mol) and dissolves t in a final volume of 0.500 L (with distilled water). She drinks the solution and proceeds to work out at 0.00-C in some mountalns where the air pressure is 650 torr. The energy she needs for her workout "galned" throvgh the reaction of the axygen (n the ais s breathing) to produce carbondionide and water viadt reaction: Cella0ee + 60 u6HAu + 6 co- a What is the M of the glecose solation? b. How many moles of oxygrn would be needed react to completely with the glucose she drank c. what is the partial pressure of oxygen in mmHg the air in these mountains is 2S0% opgen d. How many liters of air would she need to breathe to get the oxygen she needs to react completely Male oxyges la po, glucose the drank? air e. How many liters of carbon diaalde would be produced Iif all the glucose she drank reacted?

Explanation / Answer

(a)

Mass of glucose = 36.0 g.

Molar mass of glucose = 180. g/mol

Moles of glucose = mass / molar mass = 36.0 / 180. = 0.200 mol

Volume of solution = 0.500 L

Molarity of glucose of solution = moles of glucose / volume of solution

M = 0.200 / 0.500

M = 0.400 M

(b)

Balanced equation is,

C6H12O6 (s) + 6 O2 (g) -----------> 6 H2O (g) + 6 CO2 (g)

From the balanced equation above,

1 mol of glucose needs 6 mol of O2

then,

0.400 mol of glucose needs 6 * 0.400 = 2.40 mol of O2

(c)

The pressure of air = 650 torr

Partial pressure of oxygen = 650 * 25.0 / 100 = 162.5 mmHg

(d)

Pressure of oxygen = 162.5 mmHg = 162.5 / 760 = 0.214 atm

Temperature, T = 0.00 + 273.15 = 273.15 K

Using ideal gas equation,

P V = n R T

V = 2.40 * 0.0821 * 273.15 / 0.214 = 251.5 L

(e)

From the balanced equation,

6 L of O2 forms 6 L of CO2

then,

251.5 L of O2 forms 251.5 L of CO2

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