1.

Chlorine gas is prepared by reaction of `H_2SO_4` with `MnO_2` and NaCl. What volume of `Cl_2` will be produced at STP if 50 g of NaCl is taken in the reaction ?A. 1.915 LB. 22.4 LC. 11.2 LD. 9.57 L

Answer» `2NaCl+MnO_2 + 3H_2SO_4 to 2NaHSO_4 + MnSO_4 + Cl_2 + 2H_2O`
`{:("2 moles","1 mole"),("(2 x 58.5 = 117 g )", "22.4 L (STP)"):}`
117g of NaCl `-= 22.4L " of " Cl_2`
50g of NaCl`-=(22.4)/(117)xx50=9.57 L " of " Cl_2` at STP


Discussion

No Comment Found

Related InterviewSolutions