Question

A 4.55 g sample of aluminum reacted completely with a 9.62 g sample Nickel(ll) oxide. After the reaction, which reactant was still present? What mass of this reactant was present? of

Answer 1

Reaction is

2 Al + 3 NiO = Al2O3 + 3 Ni

2 mole Al react with 3 mole NiO to form one mole Al2O3 and 3 mole Ni.

4.55 gm Al = mass / molar mass = 4.55 / 26.982 = 0.1686 mole

9.62 gm NiO = 9.62 / 74.69 = 0.128 mole.

NiO is the limiting reactant and Al remains as excess.

excess reactant (Al) = 0.1686 - (0.128 * 2 / 3) = 0.0827 mole = 0.0827 *26.982 = 2.23 gm.

so 2.23 gm of reactant was present.