Question

Sodium hydroxide reacts with aluminum and water to produce hydrogen gas:2 Al(s) + 2 NaOH(aq) + 6 H2O(l) → 2 NaAl(OH)4(aq) + 3 H2(g)What mass of hydrogen gas would be formed from a reaction of 1.33 g Al and 4.25 g NaOH in water?

Answer 1

Answer: 0.147 grams

Step-by-step explanation:

To calculate the moles :

`\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}`

`\text{Moles of} Al=(1.33g)/(27g/mol)=0.049moles`

`\text{Moles of} NaOH=(4.25g)/(40g/mol)=0.106moles`

`2Al(s)+2NaOH(aq)+6H_2O(l)\rightarrow Na(Al(OH)_4(aq)+3H_2(g)`

According to stoichiometry :

2 moles of
`Al` require = 2 moles of
`NaOH/tex]</p><p>Thus 0.049 moles of [tex]Al` will require=
`(2)/(2)* 0.049=0.049moles` of
`NaOH`

Thus
`Al` is the limiting reagent as it limits the formation of product and
`NaOH` is the excess reagent.

As 2 moles of
`Al` give = 3 moles of
`H_2`

Thus 0.049 moles of
`Al` give =
`(3)/(2)* 0.049=0.0735moles` of
`H_2`

Mass of
`H_2=moles* {\text {Molar mass}}=0.0735moles* 2g/mol=0.147g`

Thus 0.147 g of
`H_2` will be produced from the given masses of both reactants.