Question

8 gm of pure calcium is treated with 50 gram of pure Hcl to give cacl2 and h2 .which is limiting reactant?​

Answer 1

Ca as a limiting reactant

Further explanation

Given

8 g Calcium

50 g HCl

Required

Limiting reactant

Solution

Reaction

Ca + 2HCl → CaCl₂ + H₂

mol Ca (Ar = 40 g/mol) :

= mass : Ar

= 8 g : 40 g/mol

= 0.2

mol HCl (MW= 36.5 g/mol) :

= mass : MW

= 50 g : 36.5 g/mol

= 1.37

Mol : coefficient reactants :

Ca = 0.2/1 = 0.2

HCl = 1.37/2 = 0.685

Ca as a limiting reactant(smaller ratio)