После пропускания 11.2 дм3 со2 через раствор кон получили 56.7 г смеси кислой и средней солей. найдите массу средней соли

1)  в  соот-ии  с  условием  произошли  2  послед-х  р-ии:   а)  koh  +  co2  =  khco3  (m = 100 г/моль)  и  б)  koh  +  khco3  =  k2co3 (m = 138 г/моль)  +  h2o  -  т.о.  koh  нах-ся  в  избытке; 2)  масса  перв-о обр-ся  khco3  сос-т  100*(11.2/22.4)  =  50  г;   в  р-ию  1б)  вст-т  х  г  khco3  и  при  этом  обр-ся  соот-но  1.38*х  г  k2co3,  что  с  учетом  массы  смеси  солей  дает  равенство  50-х+1.38*х  =  56.7,  откуда  х  =  17.632  г; 3)  т.о. масса  k2co3  сос-т  1.38*17.632  =  24.33216  г.

A)  дано:                               n(n₂)=0,75 моль         найти m(n₂            решение:   m=m*n m(n₂)= 14*2=28 m(n₂)= 28 г/моль*0,75 моль = 212 г б)           дано: n(c₆h₁₂0₆)=9*10²³               найти  m(c₆h₁₂o₆ решение: m=m*n n=n/na n=9*10²³/ 6*10²³=1,5 моль m(c₆h₁₂o₆)=12*6+1*12+16*6=180 г/моль m=180 г/моль*1,5 моль=270 г в)  дано:       n(h₂so₄)=3 кмоль   найти  m(h₂so₄ решение: m=m*n m(h₂so₄)=1*2+32+16*4=98 кг/кмоль m(h₂so₄)=  98 кг/кмоль* 3 кмоль=294 кг

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Объяснение:

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