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

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

## This Post Has 3 Comments

1. Expert says:

each  column  of the  table  is called a group. elements in a group share similar properties. groups are  read  from  top  to  bottom. metal. metalloid. nonmetal. solid. liquid. gas. fe. hg. o.  2. 3. 4. 5. 6. 7. 8. 9.  2. beryllium. 9.012. 4.  sodium. 22.990.  11. 3.  magnesium. 24.305.  12.  potassium. 39.098.  19. 4.  calcium. 40.078.  20. i guess  i

2. bessieyounger1211 says:

0.7561 g.

Explanation:

The hydrogen than can be prepared from Al according to the balanced equation:

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 (6.8 g) of Al:

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.

Now, we can get the mass of H₂ that can be prepared from 6.80 g of aluminum:

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

3. Expert says:

There is one s orbital, and there are three p orbitals, five d orbitals, and seven f orbitals. thus, the answer is b. do not confuse the number of orbitals in a subshell with the number of electrons the subshell can hold. each orbital can hold two electrons, so the capacity of an nf subshell is 14 electrons.