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Zinc sulfide reacts with oxygen according to the reaction 2ZnS(s)+3O2(g)→2ZnO(s)+2SO2(g) A reaction mixture initially contains 5.0 molZnS and 9.8 molO2. Once the reaction has occurred as completely as possible, what amount (in moles) of the excess reactant remains? Express your answer using two significant figures.

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joseaaronlara

Answer:

The answer to your question is: Excess oxygen = 2.3 mol

Explanation:

Data

ZnS = 5 mol

O2 = 9.8 mol

Excess reactant = ?

Balanced reaction

                                 2ZnS(s) + 3O2(g) → 2ZnO(s) + 2SO2(g)

MW ZnS = 65 + 32 = 97 x 2 = 194 g

MW O2 = 16 x  6 = 96 g

                           2 mol of ZnS  ------------------- 3 mol O2

 

Ratio from the reaction = 3 mol O2/ 2 mol ZnS

                                      = 1.5

Ratio from the quantities in the experiment = 9.8 mol O2 / 5 mol of ZnS

                                                                        = 1.96

Excess reactant = Oxygen because the ratio increases

                             

                                2 mol of ZnS  ------------------- 3 mol O2

                                5 mol of ZnS  -------------------  x

                                x = (5 x 3) / 2

                               x = 7.5 mol of O2

Excess Oxygen = 9.8 mol - 7.5 mol

Excess oxygen = 2.3 mol

Oseni

Going by the balanced equation of the reaction, 2.3 moles of  [tex]O_2[/tex] (g) will remain as excess at the end of the reaction

Stoichiometric calculations

From the equation of the reaction:

[tex]2ZnS(s)+3O_2(g)--- > 2ZnO(s)+2SO_2(g)[/tex]

The mole ratio of ZnS to [tex]O_2[/tex] is 2:3.

Thus, 5 moles ZnS would require 7.5 moles [tex]O_2[/tex]

This means that [tex]O_2[/tex] is in excess by: 9.8 - 7.5 = 2.3 moles

More on stoichiometric calculations can be found here: https://brainly.com/question/8062886

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