Number of grams of hydrogen than can be prepared from 6.80g of aluminum

0.7561 g.

Explanation:

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.

no. of moles of Al = mass/atomic mass = (6.8 g)/(26.98 g/mol) = 0.252 mol.

Using cross multiplication:

2.0 mol of Al produce → 3.0 mol of H₂, from stichiometry.

0.252 mol of Al need to react → ??? mol of H₂.

∴ the no. of moles of H₂ that can be prepared from 6.80 g of aluminum = (3.0 mol)(0.252 mol)/(2.0 mol) = 0.3781 mol.

mass of H₂ = (no. of moles)(molar mass) = (0.3781 mol)(2.0 g/mol) = 0.7561 g.

2b2t

explanation:

2b2t

in the analysis of systems that involve fluid flow, we frequently encounter the combination of properties u and pv.