Question

Preparation of tetraethylammonium tetrachloromanganate (II) [(C2H5)4N]2[MnCl4]

Dissolved manganese(II) chloride tetrahydrate 2.0 g in 10 mL hot absolute ethanol. Dissolved tetraethylammonium chloride manohydrate ((C2H5)4N+ 3.7 g in 10 mL hot ethanol made crystals 3.19648 g.

What is my limiting reagent? ( I guessed would be the 2.0g manganese (II) chloride tetrahydrate but I am not sure. I googled and it said the one completely used in reaction.

My yield is my product weight?

What is yield %?

Answer 1

Moles of manganese(II) chloride tetrahydrate = 2.0 g / 197.71 = 0.0100 moles

Moles of tetraethylammonium chloride manohydrate = 3.19648 g / 183.71 =0.0174 moles

2 moles of tetraethylammonium chloride manohydrate reacts with 1 mole of manganese(II) chloride tetrahydrate to give 1 mole of tetraethylammonium tetrachloromanganate (II)

For 0.0100 moles of  manganese(II) chloride tetrahydrate requires 0.0100 x2 = 0.0200 moles of tetraethylammonium chloride manohydrate .0.0174 moles .

But here we have only 0.0174 moles of tetraethylammonium chloride manohydrate .

So. the limiting reagent is tetraethylammonium chloride manohydrate .

Moles of tetraethylammonium tetrachloromanganate (II) = 0.0174/2 == 0.0087 moles

= 0.0087 x 457.25

mass = 3.978 grams