A sample of gas contains 0.1100 mol of HBr(g) and 5.00x10^-2 mol of Cl2(g) and occupies a volume of 5.85 L. The following reaction takes place: 2HBr(g) + Cl2(g) => 2HCl (g)+Br2(g)

Calculate the volume of the sample after the reaction takes place, assuming that the temperature and the pressure remain constant.

L

The volume of the sample after the reaction : V₂=5.484 L

Given

Reaction

2HBr(g) + Cl2(g) => 2HCl (g)+Br2(g)

0.1100 mol of HBr(g)

5x10⁻² mol of Cl2(g)

Required

The volume of the sample after reaction

Solution

Find limiting reactant :

0.11/2 : 0.05/1 = 0.055 : 0.05

Limiting reactant : Cl₂

mol Products based on mol Cl₂

mol products = mol HCl + mol Br₂

mol products = 2/1x0.05 + 1/1x0.05

mol products = 3 x 0.05 = 0.15

mol reactants = 0.11 + 0.05 = 0.16

From Avogadro's law,

In the same T, P, and V, the gas contains the same number of molecules  

So the ratio of gas volume will be equal to the ratio of gas moles  

 V₁/n₁=V₂/n₂

1 = reactants, 2 = products

Input the value :

5.85/0.16=V₂/0.15

V₂=5.484 L

calculate how many grams of ammonia can be produced from 4.21 grams of

n2 + 3h2 = 2nh3

my answer:

25.0g n2 x (1 mol/ 28.0gn2) x (2 mol nh3/ 1 mol n2) x (17.0g nh3/ 1 molnh3) = 30.36g n2

25.0g h2 x (1 mol h2/ 2.00g h2) x (2 mol nh3/ 3 mol h2) x (17.0g nh3/ 1 mol nh3) = 141.67g h2

lr = n2