AP Chemistry HELP Please!!!

Question

H2 Br2 -> 2HBr

A)how many grams of hydrogen bromide gas can be produced from 3.2 g of hydrogen gas and 9.5 g of bromide gas? b)How many grams of which reactant is left unreacted? C) what volume of HBr measured at STP is produced in A)?

please explain!

Answer 1

Ok, so it's been 5 years since I've taken AP Chem but I'll give this my best shot-

(1) Find the number of moles of H2 and Br2 in 3.2g and 9.5g respectively

3.2 g / (2g/mol) = 1.6mol H2

9.5 g/ (160g/mol) = 0.06 mol Br2

(2) Figure out the limiting reagent

For this problem, this is very simple since the ratio of the coefficients of H2 and Br2 is 1:1. So you can see straight away that the limiting reagent is Br2. If the coefficients hadn't been one, you would have had to divide each ones moles by its coefficient to find the limiting reagent. This means that for this reaction to go to completion, 0.06 mols H2 has to react with 0.06 mols Br2.

(3) Now that you know that Br2 is your limiting reagent, you will compare Br2 to HBr. You can see that the ratio of their coefficients is 1:2. This means that with 0.06mol Br2 you produce 0.12mol HBr. Now simply convert 0.12mol HBr to grams and you have your answer for part A.

(4) for part B, you know that your reactant in excess is your hydrogen. You also know that the amount in excess is 1.6mols-0.06mol which equals 1mol. The atomic weight of H2 is 2g/mol so you have 2g left unreacted.

(5) for part C, you need to use the relationship 1 mol of gas = 22.4L at STP. You've already computed the moles of HBr produced so now you can easily get the volume.

Hope this helps!

Answer 2

First you need to convert everything to mols

3.2 g H2 X 1mol/2g = 1.6 mols

9.5g Br2 x 1mol/160g = 0.059 mols

Bromine is the limiting reactant, therefore 2 X mols of Br gas = mols of HBr(2:1 ratio) = 0.119 mols HBr

B. Since hydrogen and bromine react in equimolar ratios, 1.6-0.059 mols of hydrogen are left = 1.54 mols X 2g/mol = 3.08g

C, 0.119 mols X 22.414L per mol = 2.667 liters

Answer 3

First you need to convert everything to mols

3.2 g H2 X 1mol/2g = 1.6 mols

9.5g Br2 x 1mol/160g = 0.059 mols

Bromine is the limiting reactant, therefore 2 X mols of Br gas = mols of HBr(2:1 ratio) = 0.119 mols HBr

B. Since hydrogen and bromine react in equimolar ratios, 1.6-0.059 mols of hydrogen are left = 1.54 mols X 2g/mol = 3.08g

C, 0.119 mols X 22.414L per mol = 2.667 liters

Answer 4

Actually

on wikipedia