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.
Answer:
First you multiply the number of atoms by the atomic mass of the element.This gives you the molar mass. Then you divide the molar mass by the number of grams in order to find the number of moles.
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