Find the mass of AlCl3 that is produced when 10.0 grams of Al2O3 react with 25.0 g of HCl according to the following equation.

Al2O3(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2O(aq) 65.4 g

help me ASAP.PLIS..

To find the mass of AlCl3 produced, we need to calculate the moles of Al2O3 and HCl separately, determine the limiting reagent, and then use stoichiometry to find the moles and mass of AlCl3 produced.

Step 1: Calculate the moles of Al2O3:
Molar mass of Al2O3 = 2(26.98 g/mol Al) + 3(16.00 g/mol O) = 101.96 g/mol

Moles of Al2O3 = Mass of Al2O3 / Molar mass of Al2O3
= 10.0 g Al2O3 / 101.96 g/mol Al2O3

Step 2: Calculate the moles of HCl:
Molar mass of HCl = 1(1.01 g/mol H) + 1(35.45 g/mol Cl) = 36.46 g/mol

Moles of HCl = Mass of HCl / Molar mass of HCl
= 25.0 g HCl / 36.46 g/mol HCl

Step 3: Determine the limiting reagent:
To determine the limiting reagent, compare the mole ratios of Al2O3 and HCl from the balanced equation. The balanced equation tells us that 1 mole of Al2O3 reacts with 6 moles of HCl.

Moles of HCl required = 6 * Moles of Al2O3

Step 4: Find the limiting reagent:
Compare the moles of HCl calculated in Step 2 and the moles of HCl required in Step 3. The reactant with the lesser number of moles is the limiting reagent.

If the moles of HCl calculated in Step 2 < moles of HCl required in Step 3, then HCl is the limiting reagent.
If the moles of HCl calculated in Step 2 > moles of HCl required in Step 3, then Al2O3 is the limiting reagent.

Step 5: Calculate the moles and mass of AlCl3 produced:
a) If HCl is the limiting reagent (using moles of HCl calculated in Step 2):
Moles of AlCl3 produced = (Moles of HCl calculated in Step 2) * (2 moles AlCl3 / 6 moles HCl)

b) If Al2O3 is the limiting reagent (using moles of Al2O3 calculated in Step 1):
Moles of AlCl3 produced = (Moles of Al2O3 calculated in Step 1) * (2 moles AlCl3 / 6 moles Al2O3)

Step 6: Convert moles of AlCl3 produced to mass:
Mass of AlCl3 = Moles of AlCl3 * Molar mass of AlCl3

Note: The molar mass of AlCl3 is given as 65.4 g/mol in the question.

I hope the steps outlined above help you in solving the problem.

To find the mass of AlCl3 that is produced, we need to calculate the theoretical yield of AlCl3 using stoichiometry.

Step 1: Find the molar masses
Molar mass of Al2O3 = 101.96 g/mol
Molar mass of HCl = 36.46 g/mol
Molar mass of AlCl3 = 2(26.98 g/mol) + 3(35.45 g/mol) = 133.34 g/mol

Step 2: Convert the given masses into moles
moles of Al2O3 = 10.0 g / 101.96 g/mol = 0.098 mol
moles of HCl = 25.0 g / 36.46 g/mol = 0.686 mol

Step 3: Determine the limiting reactant
To determine the limiting reactant, we compare the moles of each reactant to their stoichiometric coefficients in the balanced equation. From the equation, we can see that the ratio of Al2O3 to HCl is 1:6. So, if we have 0.098 moles of Al2O3, we would need 0.098 x 6 = 0.588 moles of HCl. However, we only have 0.686 moles of HCl available, which is in excess. Therefore, Al2O3 is the limiting reactant.

Step 4: Calculate the moles of AlCl3 produced
Since the balanced equation states that 1 mole of Al2O3 produces 2 moles of AlCl3, we can calculate the moles of AlCl3 produced:
moles of AlCl3 = moles of Al2O3 x (2 moles AlCl3 / 1 mole Al2O3) = 0.098 mol x 2 = 0.196 mol

Step 5: Convert moles of AlCl3 into grams
mass of AlCl3 = moles of AlCl3 x molar mass of AlCl3
mass of AlCl3 = 0.196 mol x 133.34 g/mol = 26.1024 g

Therefore, the mass of AlCl3 produced when 10.0 grams of Al2O3 reacts with 25.0 grams of HCl is approximately 26.1 grams.

This is a limiting reagent problem (LR) and you know that because amounts are given for both reactants.

Convert g Al2O3 to mols. mols = g/molar mass.

Convert g HCl to mols. same procedure.

Using the coefficients in the balanced equation, convert mols Al2O3 to mols AlCl3.
Do the same to convert mols HCl to mols AlCl3.
It is likely these to values will not agree; the correct value in LR problems is ALWAYS the smaller value and the reagent producing that value is the LR.

Now convert mols AlCl3 to g. grams = mols x molar mass AlCl3.
I don't know what the number 65.4 g means.