Suppose 6.54g of potassium bromide is dissolved in 50.ml of a 0.70 m aqueous solution of silver nitrate. calculate the final molarity of bromide anion in the solution. you can assume the volume of the solution doesn't change when the potassium bromide is dissolved in it.

[]=0.4M

Explanation:

When you add bromide to a silver nitrate solution, silver bromide can be produced. It depends of the reactants concentration.

⇄ Kps for this reaction is , this constant indicates if the reaction can happen spontaneously or not. in this case, a small value of Kps means that precipitate can be formed.

Now we need to know if our conditions are enough to form that precipitate.

=

Due to Qsp>Ksp precipitate will be formed. notice that reaction will go on until silver has been consumed completely, silver is limiting reagent.

moles of bromide remainig =total moles of bromide-moles of bromide that reacts

moles of bromide that reacts are the same number of moles from silver beacause in reaction the mole ratio is 1:1.

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2