How much KCI03 is needed to produce enough oxygen for burning of 10g of coke
Answers
2KClO3 ———- 2KCl+ 3O2
Molecular mass of KClO3= (39+ 35.5+48)=122.5U
According to the above equation
2 molecules of KClO3 gives 3 molecules of oxygen or 6 atoms of oxygen
or
245g of KClO3 gives 3(16x2)=96g of oxygen
Coke is pure carbon
C + O2 ——— CO2
1 atom of carbon needs 2 atoms of oxygen for complete burning.
i.e 12 g carbon needs 32g of oxygen
Therefore 10 g of carbon needs (32x10/ 12)=26.7g of oxygen
96g of oxygen is given by 245g of KClO3
Therefore 26.7 g of oxygen will be given by (245 x 26.7/96)=(6541.5/96)=68.14g of KClO3
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Explanation:
2KClO3 = 2KCl+ 3O2
Molecular mass of KClO3= (39+ 35.5+48)=122.5U
According to the above equation
2 molecules of KClO3 gives 3 molecules of oxygen or 6 atoms of oxygen
or
245g of KClO3 gives 3(16x2)=96g of oxygen
Coke is pure carbon
C + O2 = CO2
1 atom of carbon needs 2 atoms of oxygen for complete burning.
i.e 12 g carbon needs 32g of oxygen
Therefore 10 g of carbon needs (32x10/ 12)=26.7g of oxygen
96g of oxygen is given by 245g of KClO3
Therefore 26.7 g of oxygen will be given by (245 x 26.7/96)=(6541.5/96)=68.14g of KClO3