At STP, how many liters of H2 can be made from the complete reaction of 12.5 grams of Al? (Only enter the number, NO UNITS, and round to the tenth place, 1 decimal)
Given: 2 Al + 6 HCl → 2 AlCl3. + 3 H2
Al is about 27 grams per mol
so we have 12.5 grams * 1 mol / 27 grams = 0.463 mols of Al
for every 2 mols of Al you get 3 mols of H2
so mols of H2 = (3/2) * 0.463 = 0.694 mols H2
at STP as I recall 22.4 liters holds a mol
0.694 mols * 22.4 liters / mol =15.6 liters
To solve this problem, we need to follow a step-by-step approach:
Step 1: Convert the given mass of aluminum (Al) to moles.
Since the molar mass of Al is 26.98 g/mol, we can calculate the number of moles of Al present using the formula:
Number of moles = Mass of substance / Molar mass
Number of moles of Al = 12.5 g / 26.98 g/mol
Number of moles of Al ≈ 0.463 moles (rounded to three decimal places).
Step 2: Use the stoichiometric ratio from the balanced chemical equation to find the number of moles of hydrogen gas (H2) produced.
From the balanced equation: 2 Al + 6 HCl → 2 AlCl3 + 3 H2
We can see that 2 moles of Al produce 3 moles of H2.
So the number of moles of H2 produced is calculated as:
Number of moles of H2 = (Number of moles of Al) × (3 moles of H2 / 2 moles of Al)
Number of moles of H2 = 0.463 moles × (3/2)
Number of moles of H2 ≈ 0.695 moles (rounded to three decimal places).
Step 3: Convert the number of moles of H2 to the volume at STP (Standard Temperature and Pressure).
At STP conditions, 1 mole of any gas occupies 22.4 liters of volume.
So, the volume of H2 at STP can be calculated as:
Volume of H2 = (Number of moles of H2) × (22.4 liters/mole)
Volume of H2 = 0.695 moles × 22.4 liters/mole
Volume of H2 ≈ 15.6 liters (rounded to one decimal place).
Therefore, at STP, approximately 15.6 liters of H2 can be produced from the complete reaction of 12.5 grams of Al.