A 5.6 mol of Cl2 has a volume of 2.34 L at certain temperature and pressure. Find the new number of moles of this gas if the volume was increased to 7.2 L.
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Contents Home Bookshelves Introductory, Conceptual, and GOB Chemistry Book: Introductory Chemistry Online! (Young) 9: The Gaseous State Expand/collapse global location
9.6: Combining Stoichiometry and the Ideal Gas Laws
Last updatedAug 13, 2020
9.5: The Ideal Gas Law
9.S: The Gaseous State (Summary)
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With an understanding of the ideal gas laws, it is now possible to apply these principles to chemical stoichiometry problems. For example, zinc metal and hydrochloric acid (hydrogen chloride dissolved in water) react to form zinc (II) chloride and hydrogen gas according to the equation shown below:
2 HCl (aq) + Zn (s) → ZnCl2 (aq) + H2 (g)
Example 9.6.1 :
A sample of pure zinc with a mass of 5.98 g is reacted with excess hydrochloric acid and the (dry) hydrogen gas is collected at 25.0 ˚C and 742 mm Hg. What volume of hydrogen gas would be produced?
Solution
This is a “single state” problem, so we can solve it using the ideal gas law, PV = nRT. In order to find the volume of hydrogen gas (V), we need to know the number of moles of hydrogen that will be produced by the reaction. Our stoichiometry is simply one mole of hydrogen per mole of zinc, so we need to know the number of moles of zinc that are present in 5.98 grams of zinc metal. The temperature is given in centigrade, so we need to convert into Kelvin, and we also need to convert mm Hg into atm.
Conversions:
25.0C+273=298K(9.6.1)
(742mmHg)×(1atm760mmHg)=0.976atm(9.6.2)
(5.98gZn)×(1.00mol65.39gZn)=0.0915mol(9.6.3)
Substituting:
PV=nRT(9.6.4)
(0.976atm)×V=(0.0915mol)(0.0821Latmmol−1K−1)(298K)(9.6.5)
V=(0.0915mol)(0.0821Latmmol−1K−1)(298K)(0.976atm)