When 30.0 grams of aluminum reacts with HCl how many grams of aluminum chloride are formed? Al = HCl yields AlCl3 + H 2

the mol wt of Al is 27

30g of Al = 1.11 mol

so, we end up with 1.11 mol of AlCl3

that would be 1.11(27 + 3*35.45) = 148g

To determine the number of grams of aluminum chloride formed when 30.0 grams of aluminum reacts with HCl, we need to use stoichiometry, which involves balancing the chemical equation and using the molar ratios of the reactants and products.

Let's start by balancing the chemical equation:
2Al + 6HCl yields 2AlCl3 + 3H2

Now, we can determine the molar mass of aluminum (Al) and aluminum chloride (AlCl3):
- The molar mass of Al = 26.98 grams/mol
- The molar mass of AlCl3 = (26.98 g/mol) + 3 * (35.45 g/mol) = 133.34 grams/mol

Next, we need to convert the given mass of aluminum (30.0 grams) to moles:
moles of Al = mass / molar mass
moles of Al = 30.0 g / 26.98 g/mol ≈ 1.113 mol (rounded to three decimal places)

Now, let's use the balanced equation to determine the moles of AlCl3 formed:
From the balanced equation, we can see that the molar ratio between Al and AlCl3 is 2:2. This means that for every 2 moles of Al, we get 2 moles of AlCl3.

moles of AlCl3 = (moles of Al / 2) = 1.113 mol / 2 = 0.5565 mol (rounded to four decimal places)

Finally, we can convert the moles of AlCl3 to grams:
mass of AlCl3 = moles of AlCl3 × molar mass of AlCl3
mass of AlCl3 = 0.5565 mol × 133.34 g/mol = 74.21 grams

Therefore, when 30.0 grams of aluminum reacts with HCl, approximately 74.21 grams of aluminum chloride are formed.