To determine the number of liters of hydrogen gas produced (at STP) by the reaction of 45.0 g of aluminum metal with excess HCl, we need to follow these steps:
1. Write and balance the chemical equation for the reaction:
2 Al + 6 HCl → 2 AlCl3 + 3 H2
2. Calculate the moles of aluminum (Al) using its molar mass:
Molar mass of Al = 26.98 g/mol
Moles of Al = Mass of Al / Molar mass of Al
= 45.0 g / 26.98 g/mol
3. Determine the limiting reactant between aluminum (Al) and HCl by comparing the moles of each reactant to their stoichiometric coefficients in the balanced equation. Since HCl is in excess, aluminum is the limiting reactant.
4. Use the mole ratio from the balanced equation to find the moles of hydrogen gas (H2) produced. In this case, 2 moles of Al produce 3 moles of H2.
5. Convert moles of hydrogen gas to volume at STP (Standard Temperature and Pressure), which is 22.4 L/mol:
Volume of H2 = Moles of H2 * 22.4 L/mol
6. Calculate the volume of hydrogen gas produced at STP.
By following these steps, you can determine the volume of hydrogen gas produced at STP when 45.0 g of aluminum reacts with excess HCl.
