chemical reactions with limiting reactants?

Question

So i am confused with this question and i need a place to start the problem.

Mercury and bromine will react with each other to produce mercury (II) bromide:
Hg (l) + Br2 (l) => HgBr2 (s)
A. What is the mass of HgBr2 can be produced from the reaction of 10.0 g Hg and 9.00 g Br2?
B. What mass of which reagent is left unreacted?

You don't have to give the answers, but can someone walk me through or help me start the question? :] thanks!

Answer 1

Firstly, the reagents react in the 1:1 proportion. Now, with the help of the periodic table and a calculator we find that Hg is the limiting reagent.

mol Hg = 10.0 g / 200.59 (g/mol) = 0.0499 mol
mol Br2 = 9.00 g /159.8 (g/mol) = 0.0563 mol

Only 0.0499 of Br2 will react and you can only get this quantity of the product.

(0.0499 HgBr2) x (360.39 g/mol) = 17.98 g HgBr2

This leaves an excess of 0.0064 mol of Br2 (0.0563 - 0.0499) unreacted or 1.02 g.

Answer 2

10g of Hg is about 0.04985mol of Hg, and 9g of Br2 is about 0.05632mol of Br2. From the balanced equation above, Hg and Br2 react together in a 1:1 ratio. That means that if you have x moles of Hg, then x moles of Br2 is needed to form HgBr2.

Notice that you have an excess of Br2. Therefore, the quantity of Hg will finish first, and once it has finished, the reaction can no longer proceed. Hg is called the limiting reactant, and it determines the amount of HgBr2 that is formed.

The balanced equation above states that for every mole of Hg that is consumed, one mole of HgBr2 will be produced. Seeing that 0.04985mol of Hg will be used up, then 0.04985mol of HgBr2 will be formed. This is about 17.97g of HgBr2.

Because Br2 is in excess, some will be left over once the reaction has finished. Seeing that 0.04985mol Br2 will react with 0.04985mol Hg, then once the reaction stops, 0.00647mol Br2 (0.05632 - 0.04985) will be left over. Doing the calculations will shows that this is about 1.034g of Br2.

Answer 3

First, calculate how many moles you have of each one. Then you need to figure out which will be the limiting reagent. Since twice as much Br is needed, see if you have twice as many moles of Br as there are Hg. If there are at least twice as many moles of Br, then Hg is the limiting reagent. Since you are making 1 mole of HgBr2 for every mole of Hg the math is easy. You then figure out how many grams of HgBr2 are produced by using the molecular weight. Calculate the number of moles of Br2 are left over and then you can convert that back to grams.

If you start out with less than twice the amount of Br as Hg, then Br is the limiting reagent. You can calculate the amount of Hg left over and the amount of HgBr2 produced in a similar manner as I described above.

I hope this helps.

Answer 4

i think you must calculate how many moles for reactants and products then you can substract your products from your reactant,,
Example.
1 mole of Hg + 1 mole of Br2 ===> 1mole of HgBr2
or
x mole of Hg + x mole of Br2 ====> x mole of HgBr2