If 30.2 g of aluminum react with HCI to produce aluminum chloride and hydrogen gas, how many liters of hydrogen are produced at STP?

Question

Serenity Allen · Accepted Answer

Under #40*L# of dihydrogen gas are generated.

We must use a stoichiometric equation: #Al(s) + 3HCl(aq) rarr AlCl_3(aq) +3/2H_2(g)uarr# This equation tells us that #26.98*g# aluminum metal gives us #33.6*L# of dihydrogen gas. Why? Because #1# #mol# of gas occupies #22.4*L# at #"STP"# and #1.5*mol# were generated. #"Moles of aluminum"# #=# #(30.2*g)/(26.98*g*mol^-1)# #=# #1.12*mol#. Given the stoichiometry, #3/2xx1.12*cancel(mol)xx22.4*L*cancel(mol^-1)# are generated.

Avery Adams · Answer

undefined To calculate the volume of hydrogen gas produced at STP (Standard Temperature and Pressure) when 30.2 g of aluminum reacts with HCl, we first need to determine the moles of aluminum reacting, then use the stoichiometry of the reaction to find the moles of hydrogen gas produced, and finally convert moles to volume using the ideal gas law.

1. Calculate moles of aluminum:
   - Aluminum has a molar mass of 26.98 g/mol.
   - Moles of aluminum = mass of aluminum / molar mass of aluminum
                      = 30.2 g / 26.98 g/mol

2. Use the balanced chemical equation to find the mole ratio of aluminum to hydrogen gas:
   2Al + 6HCl → 2AlCl3 + 3H2
   From the balanced equation, 2 moles of aluminum produce 3 moles of hydrogen gas.

3. Calculate moles of hydrogen gas produced:
   - Moles of hydrogen gas = Moles of aluminum × (3 moles H2 / 2 moles Al)

4. Use the ideal gas law to find the volume of hydrogen gas at STP:
   - At STP, 1 mole of any ideal gas occupies 22.4 liters.
   - Volume of hydrogen gas = Moles of hydrogen gas × 22.4 L/mol

Calculate each step to find the final answer in liters.