Final answer:
The volume of hydrogen gas produced from the reaction of 8.88 g of gallium with excess hydrochloric acid at 300.0 K and 723 torr can be calculated using the ideal gas law after determining the moles of hydrogen. It comes out to be approximately 4.81 L.
Step-by-step explanation:
Calculating Volume of Hydrogen Gas Using the Ideal Gas Law
The question asks us to determine the volume of hydrogen gas produced at 300.0 K and 723 torr by reacting 8.88 g of gallium with excess hydrochloric acid. The chemical reaction is 2Ga(s) + 6HCl(aq) → 2GaCl₃(aq) + 3H₂(g). To solve this problem, we need to:
- Convert the gallium mass to moles using its molar mass (69.72 g/mol).
- Use stoichiometry to find the moles of hydrogen produced.
- Convert the pressure from torr to atm (723 torr is approximately 0.951 atm).
- Apply the ideal gas law (PV = nRT).
By following these steps, we get:
Molar mass of Gallium (Ga): 69.72 g/mol
- Mass of Gallium (Ga): 8.88 g
- Number of moles of Ga: 8.88 g ÷ 69.72 g/mol = 0.127 mol Ga
- Moles of H₂ produced (3 moles H₂ for every 2 moles Ga): 0.127 mol Ga × (3/2) = 0.1905 mol H₂
- Pressure in atm: 723 torr ÷ 760 torr/atm = 0.951 atm
- Volume of H₂ (V = nRT/P): 0.1905 mol × 0.08206 L atm/mol K × 300 K ÷ 0.951 atm
The calculated volume of hydrogen gas will be given by the formula V = (nRT)/P. Using the above values, the volume of hydrogen gas produced, after solving, is 4.81 L.
This is an application of the ideal gas law in stoichiometry problems to find the volume of gaseous products.