chemistry- limiting reagent?

Question

when I react 12.7 grams of Mg with excess HCl gas I expect my proceed with 50% yield. How many grams of MgCl2 should I obtain?

Answer 1

Mol(Mg) = 12.7/24 = 0.5292

Mg + 2HCl = MgCl2 + H2
Molar ratios Are;-
1 : 2 :: 1 : 1
So as molar ratios are Mg : MgCl2 :: 1 : 1
Then 0.5292 Moles (Mg) produces 0.5292mol(MgCl2)
Mr(MgCl2) =
1 x Mg = 1 x 24 = 24
2 x Cl = 2 x 35.5 = 71
24 + 71 = 95 (Mr MgCl2)
Noting that moles = mass(grams)/Mr
Mass(grams) = moles x Mr
mass = 0.5292 x 95 = 50.274 g
50.274 g is a 100% yield
25.137 g is a 50% yield.
You would expect to obtain 25.137 grams (MrCl2).

Answer 2

Mg 12.7g/24.32 g/mole = 0.522 moles of Mg
MgCl2 0.522 moles x 95.234 g/mole = 49.71 g MgCl2 at 100% yield
50% yield = 49.71/2 = 24.9 grams MgCl2