Question

ce masa de clor se obtine prin actiunea unei solutii HCL in exces aupra 153,125 g clorat de potasiu de puritate 80% ...urgenttt

Answer 1

p/100= m,pur/m,impur---> m,pur= 153,125x80/100= 1225g

ecuatia reactiei
1mol                            3mol
KClO3 +6HCl====> 3Cl2 + KCl+ 3H2o
122,5g...........................3x71g
1225g..............................m=213g Cl2.


.
M= (39gK+35,5gCl +3x16gO)/mol= 122,5g/mol