Question

Some commercial drain cleaners contain a mixture of sodium hydroxide and aluminum powder. When the mixture is poured down a clogged drain, the following reaction occurs. 2 NaOH(aq) + 2 Al(s) + 6 H2O(l) 2 NaAl(OH)4(aq) + 3 H2(g) The heat generated in this reaction helps melt away obstructions such as grease, and the hydrogen gas released stirs up the solids clogging the drain. Calculate the volume of H2 formed at 23°C and 1.00 atm if 7.01 g of Al are treated with an excess of NaOH.

Answer 1

Answer: The volume of hydrogen gas formed is 9.5 L

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 = 7.01 g

Molar mass of aluminium = 27 g/mol

Putting values in above equation, we get:

$\text{Moles of aluminium}=\frac{7.01g}{27g/mol}=0.26mol$

The given chemical equation follow:

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

As, NaOH is present in excess. It is known as excess reagent.

So, aluminium is considered as a 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, 0.26 moles of aluminium will produce = $\frac{3}[2}\times 0.26=0.39mol$ of hydrogen gas

• The equation given by ideal gas follows:

$PV=nRT$

where,

P = pressure of hydrogen gas = 1.00 atm

V = Volume of hydrogen gas = ? L

T = Temperature of hydrogen gas = $23^oC=[23+273]K=296K$

R = Gas constant = $0.0821\text{ L. atm }mol^{-1}K^{-1}$

n = number of moles of hydrogen gas = 0.39 moles

Putting values in above equation, we get:

$1.00atm\times V=0.39mol\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 273K\\\\V=\frac{0.39\times 0.0821\times 296}{1.00}=9.5L$

Hence, the volume of hydrogen gas formed is 9.5 L