Some commercial drain cleaners use a mixture of sodium hydroxide and aluminum powder.

Question

When the solid mixture is poured into the drain and dissolves, a reaction ensues that produces hydrogen gas:
2NaOH(aq) + Al(s) + 6H2O(l) → 2NaAl(OH)4(aq) + 3H2(g)

 

Determine the volume of hydrogen gas produced when 4.00 g of aluminum reacts with excess sodium hydroxide if the temperature is 40.0°C and the pressure is 101 kPa. 

Answer 1

Hello Tyler

 

We need to first use stoichiometry to convert the mass of Al to moles of H₂ gas.  From the balanced chemical reaction, you can see that 2 moles of Al reacts to form 3 moles of H₂.  Molar mass of Al = 27 g/mol.  The stoichiometry process is:

 

g Al → mol Al  → mol H₂

 

4.00 g Al x (1 mol Al / 27 g Al) x (3 mol H₂ / 2 mol Al) = 0.222 mol H₂

 

Then use the ideal gas law to solve for the volume.

 

PV = nRT

 

Value of R to use is 8.314 L kPa mol^-1 K^-1

 

T = 40.0 °C + 273.15 K = 313.2 K

 

Rearrange the ideal gas law equation to solve for V

 

V = nRT / P

 

    = 0.222 mol x 8.314 kPa L mol^-1 K^-1 x 313.2 K

       ----------------------------------------------------------------

                               101 kPa

 

    = 5.7 L