How many grams of Cl₂ gas will be obtained by the complete reaction of 31.6 gm of potassium permanganate with hydrochloric acid ? [Mole mass of `KMnO_(2) = 316 gm//"mol"`]

Options

A. 71
B. 17.75
C. 35.5
D. 142

Answer

B

Explanation

`2KMnO_(4) + 16HCl to 2KCl + 2MnCl_2+ 5Cl_(2)+ 8H_(2)O`
`("mole of " KMnO_(4))/(2) =("mole of" KCl)/(5)`
Mole of `Cl_(2)= (31.6)/(316) xx (5)/(2)= (1)/(4)` mole
Gram wt. of `Cl_(2)= (1)/(4) xx 71 = 71.75` gm