Determine the limiting reagent in Mg(s)+2HCl(aq)= MgCl2(aq)+H2(g) when Mg has .122g and HCl has 40.68g

Answer 1

See my answer to your previous question to see the procedure to use. Assuming you really meant 0.122 g of Mg and not 12.2 g or some other value, the answer below would be correct.

Mg(s) + 2HCl(aq) ==> MgCl2(aq) + H2(g)

Atomic mass Mg = 24.3 g / mol

Molar mass HCl = 36.5 g / mol

For Mg: 0.122 g x 1 mol / 24.3 g = 0.0050 mols Mg (÷1->0.005)

For HCl: 40.68 g x 1 mol / 36.5 g = 1.11 mols HCl (÷2->0.55)

0.005 is less than 0.55 so Mg is the limiting reagent.