Answer:
355.5 g.
Step-by-step explanation:
- The balanced equation for the mentioned reaction is:
2Al + 6HCl → 2AlCl₃ + 3H₂,
It is clear that 2.0 moles of Al react with 6.0 mole of HCl to produce 2.0 moles of AlCl₃ and 3.0 mole of H₂.
- Firstly, we need to calculate the no. of moles of (87.7 g) of Al:
no. of moles of Al = mass/atomic mass = (87.7 g)/(26.98 g/mol) = 3.251 mol.
Using cross multiplication:
2.0 mol of Al need to react → 6.0 mol of HCl, from stichiometry.
3.251 mol of Al need to react → ??? mol of HCl.
∴ the no. of moles of HCl needed to react with (87.7 g) 3.251 mol of Al = (3.251 mol)(6.0 mol)/(2.0 mol) = 9.752 mol.
- Now, we can get the mass of HCl needed to react with (87.7 g) 3.251 mol of Al:
mass of HCl = (no. of moles)(molar mass) = (9.752 mol)(36.46 g/mol) = 355.5 g.