Question

Calcium reacts with sulfur forming calcium sulfide. What is the theoretical yield (g) of CaS(s) that could be prepared from 7.19 g of Ca(s) and 2.67 g of sulfur(s)? Enter your answer with two decimal places. Do not type units with your answer.

Answer 1

The theoretical yield of CaS is 6.01 g.

Step-by-step explanation:

The balanced reaction is given as:

`Ca+S\rightarrow CaS`

The molar mass of Ca and S is 40.08 and 32.065 g/mol respectively.

Number of moles =
`(Mass)/(Molar Mass)`

So, 7.19 g of Ca contains
`((7.19)/(40.08))` mol of Ca or 0.179 mol of Ca

Also, 2.67 g of S contains
`((2.67)/(32.065))` mol of S or 0.0833 mol of S

According to the balanced equation:

1 mol of Ca produces 1 mol of CaS

So, 0.179 mol of Ca produces 0.179 mol of CaS

According to the balanced equation:

1 mol of S produces 1 mol of CaS

So, 0.0833 mol of S produces 0.0833 mol of CaS

As the least number of mol of CaS (product) is produced from S , therefore, S is the limiting reactant.

So, thoretically, 0.0833 mol of CaS is produced.

The molar mass of CaS is 72.143 g/mol.

So, the mass of 0.0833 mol of CaS is
`(0.0833* 72.143)` g or 6.01 g

Hence, the theoretical yield of CaS is 6.01 g.