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Carbon disulfide burns in oxygen to yield carbon dioxide and sulfur dioxide according to the following chemical equation. CS2(l) + 3O2(g) → CO2(g) + 2SO2(g)

a. If 1.00 mol CS2 reacts with 1.00 mol O2, identify the limiting reactant.
b. How many moles of excess reactant remain?
c. How many moles of each product are formed?

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Answer:

a) the limiting reactant is 02

b) There will remain 0.667 moles of CS2

c) There will be formed 0.333 moles oof CO2 and 0.667 moles of SO2

Explanation:

Step 1: Data given

Number of moles of CS2 = 1.00 mol

Number of moles of O2 = 1.00 mol

Molar mass of O2 = 32 g/mol

Molar mass of CS2 = 76.14 g/mol

Step 2: The balanced equation

CS2(l) + 3O2(g) → CO2(g) + 2SO2(g)

Step 3: Calculate the limiting reactant

For 1 mole of CS2 we need 3 moles of O2 to produce 1 mol of CO2 and 2 moles of SO2

O2 is the limiting reactant. It will completely be consumed.(1.00 mol).

CS2 is in excess. There will react 1.00/ 3 = 0.333 moles

There will remain 1.00 - 0.333 = 0.667 moles of CS2

Step 4: Calculate moles of CO2 and SO2

For 1 mole of CS2 we need 3 moles of O2 to produce 1 mol of CO2 and 2 moles of SO2

For 1.00 mol of O2 we have 1.00/3 = 0.333 moles CO2

For 1.00 mol of O2 we have 1.00 /(3/2) = 0.667 moles of SO2

In the given reaction, Oxygen is a limiting reagent while Carbon disulfide is an excess reagent.

a) The limiting reactant is Oxygen  

b) There will remain 0.667 moles of Carbon disulfid  

c) 0.333 moles of Carbon dioxide and 0.667 moles of Sulfur dioxide will  be formed.  

 

Given Here,

[tex]\bold {CS2(l) + 3O_2(g) \rightarrow CO_2(g) + 2SO_2(g)}[/tex]

1 mole of Carbon disulfide need 3 moles of Oxygen to produce 1 moles of Carbon dioxide and 2 moles of sulfur dioxide.  

Number of moles of  Carbon disulfide = 1.00 mol  

Number of moles of Oxygen = 1.00 mol    

 

Calculate the limiting reactant    

Oxygen is the limiting reactant. It will completely be consumed.(1.00 mol).  

(B) Carbon disulfide is excess reactant. There will react with

[tex]\bold { = \dfrac {1.00}{ 3 } = 0.333\ moles}[/tex]

Carbon disulfide react with 0.33 moles of Oxygen.

 

The remaining excess reagent:

[tex]\bold {1.00 - 0.333 = 0.667\ moles }[/tex]

0.667 moles of Carbon disulfide will remains.  

Calculate moles of products:

 

1 mole of Carbon disulfide need 3 moles of Oxygen to produce 1 moles of Carbon dioxide and 2 moles of sulfur dioxide.  

For 1.00 mol of oxygen =  1.00/3 = 0.333 moles Carbon dioxide.  

For 1.00 mol of Oxygen  = 1.00 /(3/2) = 0.667 moles of sulfur dioxide.

Therefore, in the given reaction, Oxygen is a limiting reagent while Carbon disulfide is an excess reagent.

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