Yellow CdS pigment is prepared by reacting ammonium sulfide with cadmium nitrate. What mass of CdS can be prepared by mixing 2.50 L of a 1.25 M Cd(NO3)2 solution with an excess of (NH4)2S?
Cd(NO3)2 + (NH4)2S ---> CdS + 2NH4NO3
Could someone give me a step by step description of how to solve this?
Sure! Here is a step-by-step process to solve this problem:
Step 1: Write down the balanced chemical equation for the reaction:
Cd(NO3)2 + (NH4)2S ---> CdS + 2NH4NO3
Step 2: Convert the given volume of Cd(NO3)2 solution into moles:
Since the concentration of Cd(NO3)2 solution is given as 1.25 M (moles per liter), we can calculate the number of moles using the formula:
moles = concentration × volume
moles = 1.25 M × 2.50 L
moles = 3.125 mol
Step 3: Determine the stoichiometry of the reaction:
From the balanced chemical equation, we can see that for every 1 mole of Cd(NO3)2, 1 mole of CdS is produced. Therefore, the moles of CdS produced will also be 3.125 mol.
Step 4: Convert moles of CdS to grams:
To find the mass of CdS, we need to multiply the moles of CdS by its molar mass. The molar mass of CdS is the sum of the atomic masses of cadmium (Cd) and sulfur (S). The atomic mass of Cd is 112.41 g/mol, and the atomic mass of S is 32.07 g/mol.
molar mass of CdS = (112.41 g/mol) + (32.07 g/mol)
molar mass of CdS = 144.48 g/mol
mass of CdS = moles of CdS × molar mass of CdS
mass of CdS = 3.125 mol × 144.48 g/mol
mass of CdS = 451.500 g
Therefore, 451.500 grams of CdS can be prepared by mixing 2.50 L of a 1.25 M Cd(NO3)2 solution with an excess of (NH4)2S.
Sure! To solve this problem, you'll need to follow a series of steps. Here's a step-by-step description of how to solve it:
Step 1: Write a balanced chemical equation.
As you mentioned, the balanced chemical equation for the reaction is:
Cd(NO3)2 + (NH4)2S ---> CdS + 2NH4NO3
Step 2: Determine the limiting reactant.
To find the limiting reactant, compare the number of moles of Cd(NO3)2 with the number of moles of (NH4)2S. To do this, you'll need to convert the given volume of 2.50 L of 1.25 M Cd(NO3)2 solution to moles.
First, calculate the number of moles of Cd(NO3)2:
Moles = Molarity x Volume
Moles of Cd(NO3)2 = 1.25 mol/L x 2.50 L = 3.125 moles
Step 3: Calculate the moles of CdS produced.
Using the stoichiometry of the balanced equation, you can see that 1 mole of Cd(NO3)2 reacts with 1 mole of CdS. Therefore, the moles of CdS produced will be the same as the moles of Cd(NO3)2.
Step 4: Convert moles of CdS to mass.
To find the mass of CdS, you'll need to know the molar mass of CdS. Looking up the atomic masses of cadmium (Cd) and sulfur (S), the molar mass of CdS is 144.47 g/mol.
Mass = Moles x Molar Mass
Mass of CdS = 3.125 moles x 144.47 g/mol = 451.5 g
Therefore, the mass of CdS that can be prepared by mixing 2.50 L of a 1.25 M Cd(NO3)2 solution with an excess of (NH4)2S is 451.5 grams.
Moles Cd(NO3)2 = M x L = ??
Using the coefficients in the balanced equation, convert moles Cd(NO3)2 to moles CdS.
Convert moles CdS to grams. g = moles x molar mass.