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:

2NaOH(aq) + 2Al(s) + 6H₂O(l) → 2NaAl(OH)₄(aq) + 3H₂(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 3.12 g of Al are treated with an excess of NaOH.

The metabolic oxidation of glucose, C₆H₁₂O₆, in our bodies produces CO₂, which is expelled from our lungs as a gas:

C₆H₁₂O₆(aq) + 6O₂(g) → 6CO₂(g) + 6H₂O(l)

Calculate the volume of CO₂ produced at body temperature (37°C) and 0.970 atm when 24.5 g of glucose is consumed in this reaction.

Calculate the volume of oxygen you would need, at 1.00 atm and 298 K, to completely oxidize 50.0 g of glucose.

Answer 1

2NaOH(aq) + 2Al(s) + 6H₂O(l) → 2NaAl(OH)₄(aq) + 3H₂(g)

As per the stoichiometry of reaction, 2moles of Al produces 3moles of H2 gas.

Molar mass Al = 27g/mol
3.12g Al = 3.12/27 = 0.1156 mol Al
This will produce 0.1156*3/2 = 0.1733 mol H2

Applying the Ideal gas equation to calculate the voume of H2, V=nRT/P

Given that, n= 0.1733, R=0.0821, T=23+273=296K, P=1atm

V=0.1733*0.0821*296/1 = 4.21 L

Thus the volume of H2 gas liberates form 3.12g of Al = 4.21 L

C₆H₁₂O₆(aq) + 6O₂(g) → 6CO₂(g) + 6H₂O(l)

As per the stiochiometry of the reaction, 1mole of glucose produces 6mol of CO2

No. of moles of glucose = (24.5gC6H12O6)(1mol/180g) =0.136 mols glucose

Now convert this to moles of CO2 by using the stoichiometric ratio of 6:1,

(0.136molsglucose)(6CO2/glucose) = 0.816mols CO2

Now use the ideal gas equation, that is, V = nRT/P, and solve:

(0.816molsCO2)(310K)(0.0821 x L x Atm) ÷ (0.970Atm x K x mol) = 21.4L CO2 produced

The volume of CO2 produced when 24.5 g of Glucose is reacted = 21.4 L

1mole of Glucose requires 6moles of O2. Thus, the no. of moles in 50.0g of Glucose = 50.0g/180g/mol = 0.277mol

The no. of moles of O2 required = 6*0.277 = 1.666mol

The volume of O2 required = V=nRT/P

Given that, n=1.666, R=0.0821L.atm, T = 298K, P=1.00atm

V = 1.666*0.0821*298/1 = 40.66 L

The volume of O2 required to completely oxidize 50.0g of Glucose = 40.66 L