4. al2o3 (s) + 6hcl (aq) → 2alcl3 (aq) + 3h20(1) find the mass of alcl3 that is produced when 10.0 grams of al2o3 react with 10.0 g of hci moar mass alqo3 102 gm imol al2o3 = 0.098 moles molar mass of hcl = 36, 5gr/mol #of moles #6l =0,274 moles mole of al2 oz 6 mol of hc 01274 x2 = 0.0913 moles alc3=13,5 6 i mass alco3 = 12, 193gm. 5. how many grams of the excess reagent in question 4 are left over?

Are produced 12,1 g of AlCl₃ and 5,33 g of Al₂O₃ are left over

Explanation:

For the reaction:

Al₂O₃ (s) + 6 HCl (aq) → 2AlCl₃ (aq) + 3H₂0(l)

10,0g of Al₂O₃ are:

10,0g ₓ = 0,0980 moles

And 10,0g of HCl are:

10,0 gₓ = 0,274 moles

For a total reaction of 0,274 moles of HCl you need:

0,274× = 0,0457 moles of Al₂O₃

Thus, limiting reactant is HCl

The grams produced of AlCl₃ are:

0,274 moles HCl × × 133 = 12,1 g of AlCl₃

The moles of Al₂O₃ that don't react are:

0,0980 moles - 0,0457 moles = 0,0523 moles

And its mass is:

0,0523 molesₓ = 5,33 g of Al₂O₃

I hope it helps!