Question

Some commercial drain cleaners contain two components: sodium hydroxide and aluminum in the form of a powder. When the mixture is poured down a clogged drain, the following redox reactions occurs: 2NaOH(aq) + 2 Al(s) + 6H2O(l) → 2NaAl(OH)4(aq) + 3 H2(g) The heat generated in this reaction helps melt away grese and the hydrogen gas released stirs up the solids clagging the drain. Calculate the volume hydrogen gas formed at 20. ºC and 750. torr if 3.12 g of Al is treated with excess NaOH.

Answer 1

Answer: The volume of hydrogen gas produced is 4.24 L

Step-by-step explanation:

To calculate the number of moles, we use the equation:

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}`

Given mass of aluminium = 3.12 g

Molar mass of aluminium = 26.98 g/mol

Putting values in above equation, we get:

`\text{Moles of aluminium}=(3.12g)/(26.98g/mol)=0.116mol`

For the given chemical reaction:

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

As, NaOH is present in excess. It is considered as an excess reagent.

Aluminium is the limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of aluminium produces 3 moles of hydrogen gas

So, 0.116 moles of aluminium will produce =
`(3)/(2)* 0.116=0.174mol` of hydrogen gas

To calculate the volume of hydrogen gas produced, we use the ideal gas equation:

`PV=nRT`

where,

P = pressure of the gas = 750 torr

V = Volume of the gas = ?

T = Temperature of the gas =
`20^oC=[20+273]K=293K`

R = Gas constant =
`62.364\text{ L. Torr }mol^(-1)K^(-1)`

n = number of moles of the gas = 0.174 mol

Putting values in above equation, we get:

`750torr* V=0.174mol* 62.364\text{ L Torr }mol^(-1)K^(-1)* 293K\\n_(mix)=\frac{0.174mol* 62.364\text{ L Torr }mol^(-1)K^(-1)* 293K}{750torr}=4.24L`

Hence, the volume of hydrogen gas produced is 4.24 L