assume that 13.5g of AL react with excess HCl according to the following balanced equation, at STP: 2Al(s) + 6HCl(aq) ---> 2AlCl3 + 3H2(g)


how mang liters of H2 at STP are produced?

To find the number of liters of H2 produced at STP, you need to use the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant (0.0821 L·atm/mol·K), and T is the temperature in Kelvin.

First, let's calculate the number of moles of H2 produced:
From the balanced equation, it is clear that for every 2 moles of Al, 3 moles of H2 are produced.
Given that you have 13.5g of Al, you can determine the number of moles of Al using its molar mass.
The molar mass of Al is 26.98 g/mol, so the number of moles of Al can be calculated as:
moles of Al = mass of Al / molar mass of Al
= 13.5 g / 26.98 g/mol
≈ 0.5002 mol

Since the stoichiometric ratio is 2:3 for Al:H2, the number of moles of H2 produced can be calculated as:
moles of H2 = (3/2) * moles of Al
= (3/2) * 0.5002 mol
≈ 0.7503 mol

Now, let's calculate the volume of H2 produced. Remember that the question asks for the volume at STP.
At STP (Standard Temperature and Pressure), the temperature (T) is 0°C or 273 K, and the pressure (P) is 1 atm.
By substituting the values into the ideal gas law equation, we can solve for the volume (V):
(1 atm) * V = (0.7503 mol) * (0.0821 L·atm/mol·K) * (273 K)
V ≈ (0.7503 * 0.0821 * 273) L
V ≈ 16.72 L

Therefore, approximately 16.72 liters of H2 are produced at STP.

mols Al = grams/molar mass = ?

Using the coefficients in the balanced equation, convert mols Al to mols H2.
Then remember that 1 mol H2 occupies 22.4 L at STP.