Question

Calcium reacts with sulfur forming calcium sulfide. what is the theoretical yield (g) of cas(s) that could be prepared from 1.6 g of ca(s) and 4.25 g of sulfur(s)? enter your answer with two decimal places. do not type units with your answer.

Answer 1

To find the theoretical yield (g) of CaS that could be prepared from 1.6 g of Ca and 4.25 g of S, we need to determine the limiting reactant. The limiting reactant is Ca because it has fewer moles than S. The theoretical yield of CaS is 2.87 g.

Step-by-step explanation:

To find the theoretical yield (g) of CaS that could be prepared from 1.6 g of Ca and 4.25 g of S, we need to determine the limiting reactant. This can be done by comparing the moles of each reactant to the stoichiometry of the balanced chemical equation. The balanced equation for the reaction is:

Ca + S → CaS

The molar mass of Ca is 40.08 g/mol and the molar mass of S is 32.06 g/mol. Using these values, we can calculate:

The moles of Ca = 1.6 g / 40.08 g/mol = 0.0399 mol

The moles of S = 4.25 g / 32.06 g/mol = 0.1325 mol

According to the stoichiometry of the balanced equation, the molar ratio between Ca and S is 1:1. Therefore, the limiting reactant is Ca because it has fewer moles than S.

To calculate the theoretical yield of CaS, we can use the moles of Ca as the basis:

Theoretical yield (g) = moles of CaS = moles of Ca x molar mass of CaS

Using the molar mass of CaS (72.14 g/mol),

Theoretical yield (g) = 0.0399 mol x 72.14 g/mol = 2.87 g

Answer 2

the balanced equation for the above reaction is;
Ca + S ---> CaS
stoichiometry of Ca to S is 1:1
Number of Ca moles - 1.6 g/ 40 g/mol = 0.04 mol
Number of S moles - 4.25 g / 32 g/mol = 0.13 mol
since Ca to S molar ratio is 1:1, there's excess amount of S present.
Ca is the limiting reactant.
stoichiometry of Ca to CaS is 1:1
Number of CaS moles to be formed - 0.04 mol
Therefore mass of CaS - 0.04 mol x 72 g/mol = 2.88 g
therefore theoretical yield is 2.88