Glucose, c 6 h 12 o 6, is a good source of food energy. when it reacts with oxygen, carbon dioxide and water are formed. how many liters of co 2 are produced when 126 g of glucose completely react with oxygen? c 6 h 12 o 6 (s) + 6o 2 (g) → 6co 2 (g) + 6h 2 o (l) + 673 kcal

94.0 L if the reaction takes place under STP.

Explanation

The molar mass of glucose C₆H₁₂O₆ is

12.01 * 6 + 1.008 * 12 + 16.00 * 16.00 = 180.16 g / mol.

126 grams of glucose will contain 126 / 180.16 = 0.69939 mol of C₆H₁₂O₆. (To avoid rounding errors, keep a couple more digits than necessary.)

6 moles of CO₂ will be produced when 1 mole of C₆H₁₂O₆ is consumed. 0.69939 moles of C₆H₁₂O₆ will give rise to 4.196 mol of CO₂.

Assuming that the reaction takes place under STP, where T = 0 °C = 273 K and P = 1 atm. Each mole of any ideal gas will occupy a volume of 22.4 liters. The 4.196 moles of CO₂ will occupy 4.196 * 22.4 = 94.0 L. (The least significant number given is 126 g, the mass of glucose. This number has three significant figures. Thus, round the result to three significant figures.)

The volume of CO₂ can be found using the ideal gas law if the condition isn't STP. For example, T = 25 °C = 297 K and P = 1.00 * 10⁵ will lead to a different volume. By the ideal gas law,

V = (n · R · T) / (P)

where

V is the volume of the gas, n is the number of moles of gas particles, R is the ideal gas constant, P is the pressure on the gas, T is the absolute temperature of the gas (in degrees Kelvin.)

R = 8.314 * 10³ L · Pa / (K · mol)

Taking T = 297 K and P = 1.00 * 10⁵ Pa,

V = (4.196 * 8.314 * 10³ * 297) / (1.00 * 10⁵) = 104 L.