Answer to Question #117837 in General Chemistry for Lori

Question

Aluminum dissolves in an aqueous solution of NaOH according to the following reaction: 2NaOH + 4Al + 4H2O -----> 2Na + 2Al2O3 + 5H2 If 500 mL of 2.20 M NaOH is reacted with 25.5 g of Al: i) Calculate the molarity of (mass) of hydrogen produced? ii) What mass of the excess reagent will remain?

Answer 1

The correct chemical equation is the following

2Al + 6NaOH +6H₂O -> 2Na₃[Al(OH)₆] + 3H₂

i) n(NaOH) = c(NaOH)*V(NaOH) = 2.20 mol/L * 0.5 L = 1.1 mol

n(Al) = m(Al)/M(Al) = 25.5 g / (26.98 g/mol) = 0.95 mol

In accordance with the chemical reaction Al is the excess reagent, thus , further calculations will be performed using chemical amount of sodium hydroxide

n(H₂) = n(NaOH)/2 = 0.55 mol; m(H₂) = 0.55 mol*2.02 g/mol = 1.11 g

ii) The amount of the excess aluminum is the following

m(Al) = n(Al)*M(Al) = (0.95 mol - 1.1 mol/3)*26.98 g/mol = 15.65 g