Question

Preparation of tetraethylammonium tetrachloromanganate (II) [(C2H5)4N]2[MnCl4] Dissolved manganese(II) chloride tetrahydrate 2.0 g in 10 mL hot...

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 %?

Homework Answers

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

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT