Question

Oblicz skaład procentowy następujacych związków

a)FeSO3

b)CaSiO3

c)AIBO3

d)K2CO3

Answer 1

a)FeSO3 = 56u+32u+48u=136u

%Fe=  136u -100%

56u - x%

136u*x=56u*100

136u=5600/ :136

x=41,17%

%S= 136u - 100%

32u - x%

136u*x=32u*100

136u=3200/ :136

x=23,52%

 

%O=100%-64,69%=35,31%

 

b) CaSiO3= 40u+28u+48u=116u

%Ca= 116u - 100%

40u - x%

116u*x%=40u*100%

116u=4000/ :116

x=34,48%

 

%Si = 116u-100%

28u-x%

116u*x%=28u*100%

116u=2800/ :116

x=24,13%

 

%O= 100%-58,61%=41,39%

 

c) AlBO3=27u+11u+48u=86u

 

%Al= 86u-100%

27u-x%

86u*x%=27u*100%

86u=2700/ :86

x=31,39%

 

%B= 86u-100%

11u-x%

86u*x%=11u*100%

86u=1100/ :86

x=12,79%

 

%O= 100%-44,18%=55,82%

 

d) K2CO3= 78u+12u+48u=138u

 

%K= 138u-100%

78u-x%

138u*x%=78u*100%

138u=7800/ :138

x=56,52%

 

%C= 138u-100%

12u-x%

 

138u*x%=12u*100%

138u=1200/:138

x=8,69%

 

%O=100%-65,21%=34,79%

Answer 2

a)55u+32u+3*16u=135

    135u-100%

     55u -x

   x=100%*55u/135u

   x=40,7 %mFE 

   135u-100%

     32u-x

x=3200/150

x=21,3 %mS

   100%-(21,3+40,7)=

    100-62=28%mO

b)40+28+2*16=100u

   100u-100%

        40u-x

      x=100%*40u/100u

      x=40%mca

      100u-100%

        28u-x

     x=100%*28u/100u

     x=28%%msi 

     100%-(40+28)=32%mO

      c)

         27u+14u+16*3=89

          89-100

          27-x

       x=2700/89   

       x=30,3%mAl

       89-100

        14-x

          x=1400/89

           x= 15,7%mN

           100%-46=54%mo

        d)39*2+12+16*3=138u

         138u-100%

           78U-X

         X=7800/138

         X=56,5%MK

           138-100%

            12-X

           X=1200/138 

           X8,7%MC

           100%-56,5+8,7=34,8%MO