Question

Report Sheet - Lab 30A-3 Q.5 Compare the calculations of the moles of Mg that should have reacted with the acid (3.3) with the observations of Mg remaining in the bottom of the flask in A.2, A.3 and A.4. Based on the calculations, which flasks should have had some of magnesium remaining? Did this match your observations? Q.6 For the reaction of magnesium and HCI. if you react 25 grams of magnesium with 37 grams of HCl, which reactant is the limiting reactant and how much MgCl, will be produced by the reaction?

Answer 1

Given reaction :-

Mg + 2 HCl -> MgCl2 + H2

Mass of Mg = 25g

Mole of Mg = given mass/ Mw

= 25g/24.3g mol-1

= 1.0288mol

Mass of HCl = 37 g

Mole of HCl = 37g/36.46g mol-1

= 1.0148mol

In order to find limiting reactant, we divide number of moles of reactants with their stoichiometric coefficient. Whichever reactant has lower value will be limiting reactant and products will form according to it.

For Mg, 1.0288/1 = 1.0288

For HCl, 1.0148/2 = 0.5074

So, HCl has lower value. It is the limiting reactant and products will form according to it.

Now,

2 mol of HCl produces 1 mol of MgCl2

So, 1 mol of HCl produces 1/2 mol of MgCl2

So, 1.0148 mol of HCl produces 1.0148/2 mol. of MgCl2.

Mole of MgCl2 produced = 0.5074 mol

So, mass of MgCl2 produced = mol X Mw

= 0.5074 mol X 95.2 g.mol-1

= 48.3 g