Question

The following reaction takes place when some drain cleaners are added to water. What mass of hydrogen gas will be produced from 4.76 g of aluminum and 3.23 g of NaOH? 2Al(s) + 2 NaOH (s) + 6 H2O (l) ----> 2NaAl(OH)4 (aq) + 3 H2 (g)

Answer 1

Number of moles of Al = 4.76 g / 26.9815 g/mol = 0.176 mol

Number of moles of NaOH = 3.23 g / 39.997 g/mol = 0.0808 mol

From the balanced equation we can say that

2 mole of Al requires 2 mole of NaOH so

0.176 mole of Al will require

= 0.176 mole of Al *( 2 mole of NaOH / 2 mole of Al )

= 0.176 mole of NaOH

But we have 0.0808 mole of NaOH which is in short so NaOH is limiting reactant

From the balanced equation we can say that

2 mole of NaOH produces 3 mole of H2 so

0.0808 mole of NaOH will produce

= 0.0808 mole of NaOH *(3 mole of H2 / 2 mole of NaOH )

= 0.121 mole of H2

mass of 1 mole of H2 = 2.016 g

so the mass of 0.121 mole of H2 = 0.244 g

Therefore, the mass of H2 produced would be 0.244 g