Question

Find the amount of potassium chlorate subjected to thermal decomposition to give oxygen which was sufficient for the combustion of 120 g of ethane ($C_2{H}_6$).

• A. 9.3 moles
• B. 3.1 moles
• C. 6.2 moles
• D. 9.2 moles

Answer 1

The correct option is A 9.3 moles

$2C_2{H}_6+{70}_2=4C{O}_2+6H_{2}O$

2 mole of $C_2{H}_6$ require 7mole of oxygen

we have $C_2{H}_6$ moles = 120g / 30(g/mol) = 4 mol

for 4 mol we require 14 mol oxygen

$2KCl{O}_3\to 2KCl+3O_2$

1 mole $KCl{O}_3$ Gives = 1.5 mol $O_2$

for 14 mol $O_2$ We need : $\frac{1}{1.5}\times 14$ = 9.3 mol of $KCl{O}_3$

ans is A