what mass of CdS is produced if 4.55g of cadmium reacts with 2.00g of sulfur?
To find the mass of CdS produced, we need to determine the limiting reactant between cadmium (Cd) and sulfur (S). The limiting reactant is the one that is completely consumed and determines the maximum amount of product that can be formed.
To determine the limiting reactant, we compare the amount of each reactant to the stoichiometric ratio in the balanced chemical equation. The balanced equation for the reaction between cadmium and sulfur to form cadmium sulfide (CdS) is:
Cd + S -> CdS
The molar mass of Cd is 112.41 g/mol, and the molar mass of S is 32.06 g/mol.
Step 1: Calculate the number of moles of each reactant.
Number of moles of Cd = mass of Cd / molar mass of Cd
Number of moles of Cd = 4.55 g / 112.41 g/mol = 0.0405 mol
Number of moles of S = mass of S / molar mass of S
Number of moles of S = 2.00 g / 32.06 g/mol = 0.0624 mol
Step 2: Determine the limiting reactant.
To compare the amount of Cd and S, we need to consider the ratio in the balanced equation. From the balanced equation, we can see that the stoichiometric ratio is 1:1 between Cd and S.
Therefore, Cd is the limiting reactant since it is not in excess. There is less Cd available compared to the stoichiometric ratio.
Step 3: Calculate the mass of CdS produced.
Since Cd is the limiting reactant, the amount of CdS produced is equal to the amount of Cd consumed.
Number of moles of CdS = moles of limiting reactant (Cd) = 0.0405 mol
Mass of CdS = number of moles of CdS × molar mass of CdS
Mass of CdS = 0.0405 mol × (molar mass of Cd + molar mass of S)
Mass of CdS = 0.0405 mol × (112.41 g/mol + 32.06 g/mol)
Mass of CdS = 7.79 g
Therefore, the mass of CdS produced is 7.79 grams.