Answer to Question #201576 in Math for rama

Question #201576

In a chemical reaction of Propane and oxygen gives the carbon dioxide and water as

given below

C3H3 + O2 -! CO2 + H2O

Using Gauss Seidel method Find the numbers of atoms of C, H, O.


1
Expert's answer
2021-06-03T13:22:59-0400

Write the unbalanced combustion reaction as follows


C3H3(g)+O2(g)CO2(g)+H2O(g)C_3H_3(g)+O_2(g)\to CO_2(g)+H_2O (g)

Balance the equation as follows

Consider HH

The least common multiplier of 33 in left side and 22 in right side is 6.6. Then


2C3H3(g)+O2(g)CO2(g)+3H2O(g)2C_3H_3(g)+O_2(g)\to CO_2(g)+3H_2O (g)


Consider CC

The least common multiplier of 2(3)2(3) in left side and 11 in right side is 6.6. Then


2C3H3(g)+O2(g)6CO2(g)+3H2O(g)2C_3H_3(g)+O_2(g)\to 6CO_2(g)+3H_2O (g)


Check OO

We have even number of atoms of OO in left side and odd number of atoms of OO in right side. Multiple both sides by 22

4C3H3(g)+2O2(g)12CO2(g)+6H2O(g)4C_3H_3(g)+2O_2(g)\to 12CO_2(g)+6H_2O (g)

We have 22 atoms of OO in left side and 12(2)+6=3012(2)+6=30 atoms of OO in right side. Then


4C3H3(g)+15O2(g)=12CO2(g)+6H2O(g)4C_3H_3(g)+15O_2(g)= 12CO_2(g)+6H_2O (g)

12 atoms of CC


12 atoms of HH


30 atoms of OO



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