Question

what is the maximum no. of grams of cu 2s that can be formed when 80 g of cu reacts with 25 g of s

Answer 1

Answer:

100.27 g of $Cu_{2} S$ is formed when 80 g of Cu reacts with 25 g of $S_{8}$.

Explanation:

The balanced chemical equation for the reaction can be written as:

$16Cu + S_{8}$ → $8Cu_{2} S$

Given mass of Cu = 80 g

Molar mass of Cu = 63.55 g/mol

Number of moles of Cu = Given mass ÷ molar mass

                                      = $\frac{80}{63.55}$

                                      = 1.26 moles

Given mass of  $S_{8}$= 25 g

Molar mass of $S_{8}$ = 256 g/mol

Number of moles of  $S_{8}$ = Given mass ÷ molar mass

                                       =  $\frac{25}{256}$

                                       = 0.098 moles

1 mole of $S_{8}$ reacts with 16 moles of Cu

∴0.098 moles of $S_{8}$ should react with 0.098 × 16 = 1.56 moles

But we have only 1.26 moles of Cu.

∴Cu is the limiting reagent.

16 moles of Cu gives 8 moles of $Cu_{2} S$

⇒1.26 moles of Cu gives $\frac{8}{16}$ × 1.26 = 0.63 moles of $Cu_{2} S$

Molar mass of $Cu_{2} S$ = 159.16 g/mol

∴Mass of $Cu_{2} S$ obtained = 0.63 × 159.16

                                         = 100.27 g