Question

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?

Answer 1

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

Step-by-step explanation:

For the reaction:

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

10,0g of Al₂O₃ are:

10,0g ₓ
`(1mol)/(102g)` = 0,0980 moles

And 10,0g of HCl are:

10,0 gₓ
`(1mol)/(36,5g)` = 0,274 moles

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

0,274×
`(1molesAl_(2)O_3)/(6 mole HCl)` = 0,0457 moles of Al₂O₃

Thus, limiting reactant is HCl

The grams produced of AlCl₃ are:

0,274 moles HCl ×
`(2 moles AlCl_(3))/(6 moles HCl)` × 133
`(g)/(mol)` = 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
`(102g)/(1mol)` = 5,33 g of Al₂O₃

I hope it helps!