What volume (in mL) of 0.397 mol/L sodium hydroxide, NaOH (aq), is needed to completely neutralize 54.5 mL of 0.349 mol/L sulfuric acid, H2SO4(aq)?

Question

Some commercial drain cleaners use a mixture of sodium hydroxide and aluminum powder. When the solid mixture is poured into the drain and dissolves, a reaction ensues that produces hydrogen gas:

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

Determine the temperature (in ^°C) of hydrogen gas produced when 37.29 g of aluminum reacts with excess sodium hydroxide and water if the pressure is 117.07 kPa and the volume is 57.27 L. Provide your answer with TWO decimals.

Answer 1

Volume of NaOH needed to neutralize H2SO4:

H2SO4 + 2 NaOH ==> Na2SO4 + 2H2O ... balanced equation

mols H2SO4 present = 54.5 mls x 1 L / 1000 mls x 0.349 mol / L = 0.0190 mols H2SO4

mols NaOH needed = 0.0190 mols H2SO4 x 2 mols NaOH/mol H2SO4 = 0.0380 mols NaOH needed

volume NaOH needed: (0.397 mol/L)(x L) 0.0380 mols and x = 0.09582 L = 95.8 mls (3 sig. figs.)

Temperature of H2 gas:

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

mols Al present = 37.29 g Al x 1 mol Al/26.98 g = 1.382 mols Al

mols H2 gas produced = 1.382 mols Al x 3 mols H2 / 2 mols Al = 2.073 mols H2 gas produced

Using the ideal gas law, we can solve for temperature...

PV = nRT

P = pressure = 117.07 kPa x 1 atm/101.325 kPa = 1.155 atm

V = volume = 57.27 L

n = moles of gas = 2.073 mols H₂

R = gas constant = 0.0821 Latm/Kmol

T = temperature in Kelvin = ?

Solving for T, we have ...

T = PV/nR = (1.155 atm)(57.27 L) / (2.073 mols)(0.0821 Latm/Kmol)

T = 388.66K

Converting this to ºC we have...

388.66 - 273.15 = 115.5ºC = T