consider the following reaction AgNO3+KBr=AgBr+KNO3
calculate the amount of presipitate in gram obtained when 0.25 mole of AgNO3 is trheated with excess of aqueous KBr solution
you get 0.25 moles of AgBr, so what is its mass?
since their is 0.25 moles of AgNO3 hence you the mass = 42.5g and similarly mass of kbr =42.5 g or 0.35 moles by GM/MM = NO OF MOLES
hence by this we can get that their is 0.25 moles of AgBr thier for its mass is 47 g by applying law of conservation of mass
AgNO3 + KBr → AgBr ↓ + KNO3
1M 1M
0.25M 0.25M
0.25M of AgBr=108+80=188/4=47
Ans: 47
Well, let me do some calculations (while juggling some rubber chickens) for you.
According to the balanced chemical equation, 1 mole of AgNO3 reacts with 1 mole of KBr to produce 1 mole of AgBr.
Since you have 0.25 moles of AgNO3, that means you will react with an equal amount of 0.25 moles of KBr.
Now, the molar mass of AgBr is approximately 187.77 g/mol. So, we can calculate the mass of AgBr precipitate using the equation:
Mass of AgBr = Number of moles of AgBr x Molar mass of AgBr
Since we know that 1 mole of AgNO3 reacts with 1 mole of AgBr, we have 0.25 moles of AgBr.
Calculating:
Mass of AgBr = 0.25 moles x 187.77 g/mol
The mass of AgBr precipitate obtained is approximately 46.94 grams.
So, after heating 0.25 moles of AgNO3 with excess aqueous KBr solution, you will get approximately 46.94 grams of AgBr precipitate. But be careful, you might end up with some clowns popping out instead!
To calculate the amount of precipitate produced when 0.25 mole of AgNO3 reacts with an excess of aqueous KBr solution, we need to determine the limiting reactant first. The limiting reactant is the one that is completely consumed, determining the maximum amount of product that can be formed.
In this reaction, the molar ratio between AgNO3 and AgBr is 1:1. Thus, if 1 mole of AgNO3 reacts, 1 mole of AgBr is obtained. Therefore, the moles of AgBr formed will be equal to the moles of AgNO3.
Given that there are 0.25 moles of AgNO3, we can conclude that there will be 0.25 moles of AgBr precipitate formed.
To convert moles to grams, we need to know the molar mass of AgBr. The molar mass of AgBr can be calculated by adding the atomic masses of silver (Ag) and bromine (Br):
Ag: 107.87 g/mol
Br: 79.90 g/mol
Molar mass of AgBr = (107.87 g/mol) + (79.90 g/mol) = 187.77 g/mol
Now we can calculate the amount of precipitate in grams by multiplying the moles of AgBr by its molar mass:
Mass of AgBr = moles of AgBr × molar mass of AgBr
Mass of AgBr = 0.25 moles × 187.77 g/mol
Using this equation, the mass of AgBr precipitate obtained when 0.25 moles of AgNO3 reacts with excess KBr can be calculated.