2 gm of a mixture of `CO and CO_(2)` on reaction with excess of `I_(2)O_(5)` produced 2.54 gm of `I_(2)`. What will be the mass `%` of `CO_(2)` in the original mixture ?

Options

A. 60
B. 30
C. 70
D. 35

Answer

B

Explanation

`5CO+I_(2)O_(5) to 5CO_(2)+I_(2)`
1 mol of `I_(2)-= "1 mole of" I_(2)O_(5)-= "5mol of CO"`
Hence, mol of `CO=5 xx (2.54)/(254)=0.05`
Mass of CO `=0.05 xx 28=1.4 g`
Mass of `CO_(2)` =2 - 1.4 = 0.6 g
Mass % of `CO_(2)=(0.6)/(2) xx 100=30%`