Calculation of Precipitate in Grams

To calculate the precipitate obtained when 0.25 moles of AgNO3 reacts with an excess of aqueous KBr solution, we need to determine the limiting reactant and use stoichiometry to find the amount of precipitate formed.

Step 1: Determine the Limiting Reactant
The limiting reactant is the one that is completely consumed first, thus determining the maximum amount of product formed. To find the limiting reactant, we compare the moles of AgNO3 and KBr.

Given:
Moles of AgNO3 = 0.25 moles

The balanced chemical equation for the reaction is:
AgNO3 + KBr → AgBr + KNO3

From the equation, we can see that the stoichiometric ratio between AgNO3 and AgBr is 1:1. This means that every mole of AgNO3 will produce one mole of AgBr.

Therefore, the moles of AgBr formed will also be 0.25 moles.

Step 2: Convert Moles of AgBr to Grams
To convert moles of AgBr to grams, we need to know the molar mass of AgBr.

The molar mass of AgBr can be calculated as follows:
Ag = 107.87 g/mol
Br = 79.90 g/mol

Molar mass of AgBr = Ag + Br = 107.87 + 79.90 = 187.77 g/mol

Now, we can calculate the mass of AgBr formed using the moles of AgBr:

Mass of AgBr = Moles of AgBr × Molar mass of AgBr
= 0.25 moles × 187.77 g/mol
= 46.94 grams

Therefore, when 0.25 moles of AgNO3 reacts with an excess of aqueous KBr solution, the mass of precipitate (AgBr) obtained will be 46.94 grams.

Summary:
- The limiting reactant in this reaction is AgNO3.
- The moles of AgBr formed will also be 0.25 moles.
- The molar mass of AgBr is 187.77 g/mol.
- The mass of AgBr precipitate obtained will be 46.94 grams.

47 gm of AgBr will precipitate out