Question

calculate the amount of of Kclo3 which gives enough oxygen for the combustion of 32g of (ch)4



please answer ​

Answer 1

VERIFIED ANSWER

Answer:

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