Ta có `1đvC=1,6605xx10^{-24}(g)`

`a)`

`PTK_{Fe_3O_4}=56xx3+16xx4=232(đvC)`

`=1,6605xx10^{-24}xx232=3,85236.10^{-22}(g)`

`b)`

`PTK_{3BaCO_3}=3.(137+12+16.3)=591(đvC)`

`=1,6605.10^{-24}.591=9,813555.10^{-22}(g)`

`c)`

`PTK_{Fe_2(SO_4)_3}=56.2+(32+16.4).3=400(đvC)`

`=1,6605.10^{-24}.400=6,642.10^{-22}(g)`

`d)`

`PTK_{5Ca_3(PO_4)_2}=5.[40.3+(31+16.4).2]=1550(đvC)`

`=1550.1,6605.10^{-24}=2,573775.10^{-21}(g)`

`e)`

`PTK_{Na_2CO_3.10H_2O}=23.2+12+16.3+10.18=286(đvC)`

`=286.1,6605.10^{-24}=4,74903.10^{-22}(g)`

`g)`

`PTK_{3H_2SO_4.5SO_3}=3.[2+32+16.4+5.(32+16.3)]=1494(đvC)`

`=1494.1,6605.10^{-24}=2,480787.10^{-22}(g)`

Đáp án:
a)
`PTK_{Fe_3O_4}=56 xx 3 + 16 xx 4= 232 (đvC)`
`=> m_{Fe_3O_4}=232 xx 1,6605.10^{-23}= 385,236.10^{-23} (g)`
b)
`PTK_{3BaCO_3}=3 xx (137 + 12 + 16 xx 3)= 591 (đvC)`
`=> m_{3BaCO_3}=591 xx 1,6605.10^{-23}= 981,3555.10^{-23} (g)`
c)
`PTK_{Fe_2(SO_4)_3}=56 xx 2 + (32 + 16 xx 4) xx 3= 400 (đvC)`
`=> m_{Fe_2(SO_4)_3}=400 xx 1,6605.10^{-23}= 664,2.10^{-23} (g)`
d)
`PTK_{5Ca_3(PO_4)_2}=5 xx (40 xx 3 + 31 xx 2 + 16 xx 4 xx 2)= 1550 (đvC)`
`=> m_{5Ca_3(PO_4)_2}=1550 xx 1,6605.10^{-23}= 2573,775.10^{-23} (g)`
e)
`PTK_{Na_2CO_3.10H_2O}=23 xx 2 + 12 + 16 xx 3 + 10 xx (1 xx 2 + 16)= 286 (đvC)`

`=> m_{Na_2CO_3.10H_2O}=286 xx 1,6605.10^{-23}= 474,903.10^{-23}`

g)

`PTK_{3H_2SO_4.5SO_3}=3 xx (1 xx 2 + 32 + 16 xx 4 + 5 xx 80)= 1494 (đvC)`

`=> m_{3H_2SO_4.5SO_3}=1494 xx 1,6605.10^{-23}= 2480,787.10^{-23}`