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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.

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Answer:

The theoretical yield of CaS is 6.01 g.

Explanation:

The balanced reaction is given as:

[tex]Ca+S\rightarrow CaS[/tex]

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

Number of moles = [tex]\frac{Mass}{Molar Mass}[/tex]

So, 7.19 g of Ca contains [tex](\frac{7.19}{40.08})[/tex] mol of Ca or 0.179 mol of Ca

Also, 2.67 g of S contains [tex](\frac{2.67}{32.065})[/tex] 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 [tex](0.0833\times 72.143)[/tex] g or 6.01 g

Hence, the theoretical yield of CaS is 6.01 g.