Calculate the mass of silver bromide produced from 22.5 g of silver nitrate in the following rxn: 2AgNO3(aq) + MgBr2(aq)-->2AgBr(s)+ Mg(NO3)2 (aq)
To calculate the mass of silver bromide produced, we need to determine the molar mass of silver bromide (AgBr) and the stoichiometry of the reaction.
Step 1: Calculate the molar mass of silver bromide (AgBr).
The molar mass of silver (Ag) is 107.87 g/mol, and the molar mass of bromine (Br) is 79.90 g/mol. Adding these two values gives us the molar mass of AgBr:
Molar mass of AgBr = (107.87 g/mol) + (79.90 g/mol) = 187.77 g/mol
Step 2: Determine the stoichiometry of the reaction.
From the balanced equation, we can see that 2 moles of AgNO3 react with 1 mole of MgBr2 to produce 2 moles of AgBr. This means that for every 2 moles of AgNO3, we will get 2 moles of AgBr.
Step 3: Convert the given mass of AgNO3 to moles.
Given mass of AgNO3 = 22.5 g
To convert grams to moles, we need to divide the mass by the molar mass:
Moles of AgNO3 = (22.5 g) / (169.87 g/mol) = 0.1325 mol (rounded to 4 decimal places)
Step 4: Calculate the moles of AgBr produced.
Since we have a 1:1 stoichiometric ratio between AgNO3 and AgBr, we can conclude that 0.1325 moles of AgNO3 will produce 0.1325 moles of AgBr.
Step 5: Calculate the mass of AgBr produced.
To calculate the mass, we use the equation:
Mass of AgBr = Moles of AgBr × Molar mass of AgBr
Mass of AgBr = (0.1325 mol) × (187.77 g/mol) = 24.85 g (rounded to 2 decimal places)
Therefore, the mass of silver bromide produced from 22.5 g of silver nitrate is 24.85 g.
To calculate the mass of silver bromide (AgBr) produced, we need to determine the limiting reagent first. The limiting reagent is the reactant that is completely consumed, thereby determining the maximum amount of product that can be formed.
1. Calculate the molar mass of silver nitrate (AgNO3):
AgNO3: Ag (107.87 g/mol) + N (14.01 g/mol) + 3O (3x16.00 g/mol) = 169.87 g/mol
2. Convert the mass of silver nitrate to moles:
Moles of AgNO3 = mass of AgNO3 / molar mass of AgNO3
Moles of AgNO3 = 22.5 g / 169.87 g/mol ≈ 0.1325 mol
3. Use the stoichiometry of the balanced equation to determine the moles of AgBr produced from AgNO3:
From the balanced equation, we have a 1:2 ratio between AgNO3 and AgBr:
Moles of AgBr = 0.1325 mol × (2 mol AgBr / 2 mol AgNO3) = 0.1325 mol
4. Calculate the mass of AgBr produced from the moles calculated:
Mass of AgBr = Moles of AgBr × molar mass of AgBr
Molar mass of AgBr: Ag (107.87 g/mol) + Br (79.90 g/mol) = 187.77 g/mol
Mass of AgBr = 0.1325 mol × 187.77 g/mol ≈ 24.87 g
Therefore, the mass of silver bromide produced from 22.5 g of silver nitrate is approximately 24.87 g.