100 ml of 0.2 M ammonium sulfate are reacted with 300 ml of 0.1M cadmium chloride . a yellow precipitate is formed. what is the weight of the precipitate produced.
I think you have made a typo. These two substances don't react. (NH4)2S I might believe.
Nope the question is exactly like this!!!
Then answer it that there is no reaction. If that is exactly how the problem is stated then the author made a typo.
To determine the weight of the precipitate produced, we need to determine the limiting reactant and calculate the amount of precipitate formed using the stoichiometry of the reaction.
First, let's write the balanced chemical equation for the reaction between ammonium sulfate (NH4)2SO4 and cadmium chloride CdCl2:
(NH4)2SO4 + CdCl2 → CdSO4 + 2NH4Cl
To find the limiting reactant, we compare the number of moles of each reactant.
Moles of (NH4)2SO4 = volume of (NH4)2SO4 solution (in L) × concentration of (NH4)2SO4 (in mol/L)
Moles of CdCl2 = volume of CdCl2 solution (in L) × concentration of CdCl2 (in mol/L)
Calculate the moles of (NH4)2SO4:
Moles of (NH4)2SO4 = 0.100 L × 0.2 mol/L = 0.02 mol
Calculate the moles of CdCl2:
Moles of CdCl2 = 0.300 L × 0.1 mol/L = 0.03 mol
According to the balanced equation, the stoichiometric ratio between (NH4)2SO4 and CdSO4 is 1:1. This means that for every 1 mole of (NH4)2SO4 reacted, 1 mole of CdSO4 is formed.
Since (NH4)2SO4 is the limiting reactant (0.02 mol) and the stoichiometric ratio is 1:1, we can conclude that 0.02 mol of CdSO4 will be formed.
To calculate the weight of the precipitate, we need to know the molar mass of CdSO4. The molar mass of CdSO4 is:
Cd = 112.41 g/mol
S = 32.07 g/mol
Oxygen (4 × 16.00) = 64.00 g/mol
Total molar mass = 112.41 + 32.07 + 64.00 = 208.48 g/mol
Now, we can calculate the weight of the precipitate:
Weight of CdSO4 = moles of CdSO4 × molar mass of CdSO4
Weight of CdSO4 = 0.02 mol × 208.48 g/mol
Therefore, the weight of the precipitate produced is 4.17 grams (rounded to two decimal places).