94.0 L if the reaction takes place under STP.
<h3>Explanation</h3>
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<em>,</em>
- 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.
