Final answer:
To calculate the volume of H₂ gas produced, calculate the moles of lithium initially present, and then apply the ideal gas law using the pressure and temperature provided, converting the final volume to milliliters.
Step-by-step explanation:
To determine the volume of H₂ gas formed when 0.0108 g of Lithium metal is reacted in 5.00 mL of 6M HCl, we first have to use the stoichiometry of the reaction between lithium and hydrochloric acid to find the number of moles of hydrogen gas produced. Using the molar mass of lithium (6.94 g/mol), we calculate the moles of lithium, and hence the moles of hydrogen gas since the reaction produces one mole of H₂ for every mole of lithium. We then apply the ideal gas law (PV = nRT) using the given values of pressure (0.957 atm) and temperature (converted to Kelvin, which is 21.5°C + 273.15 = 294.65 K).
The calculation will follow this format:
- Calculate moles of Li: (0.0108 g Li) / (6.94 g/mol Li) = number of moles of Li
- Since the reaction is 2Li(s) + 2HCl(aq) → 2LiCl(aq) + H₂(g), the moles of H₂ are the same as those of Li.
- Volume of H₂ (V) = (n)(R)(T) / P
- Insert the values: moles of H₂ × 0.08206 L atm mol⁻¹ K⁻¹ × 294.65 K / 0.957 atm = volume in liters
- Convert the volume from liters to milliliters by multiplying by 1,000.
This approach will yield the volume of H₂ gas in milliliters. Remember to ensure all units are consistent when applying the ideal gas law.