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1.

`a)`

`n_{K_3PO_4}={250}/{212}={125}/{106}(mol)`

`200ml=0,2l`

`->C_{M\ K_3PO_4}={{125}/{106}}/{0,2}\approx 5,9M`

`b)`

`n_{CH_4}={12,06.10^{22}}/{6,022.10^{23}}\approx 0,2(mol)`

`800ml=0,8l`

`->C_{M\ CH_4}={0,2}/{0,8}=0,25M`

`c)`

`n_{Fe_2(SO_4)_3}={250.6\%}/{400}=0,0375(mol)`

`400ml=0,4l`

`->C_{M\ Fe_2(SO_4)_3}={0,0375}/{0,4}=0,09375M`

2.

`a)`

`350ml=0,35l`

`->n_{ZnCl_2}=0,35.2=0,7(mol)`

`->C\%_{ZnCl_2}={0,7.136}/{600}.100\%\approx 15,867\%`

`b)`

`n_{Ba(OH)_2}=4.0,3=1,2(mol)`

`->C\%_{Ba(OH)_2}={1,2.171}/{250}.100\%=82,08\%`

3.

`a)`

`20ml=0,02l`

`->n_{Ca_3(PO_4)_2}=0,02.3=0,06(mol)`

`->m_{Ca_3(PO_4)_2}=0,06.310=18,6(g)`

`b)`

`n_{Al_2(SO_4)_3}=5.4=20(mol)`

`->m_{Al_2(SO_4)_3}=20.342=6840(g)`

1)

a)`n_{K_3PO_4}` = `{250}/{212}` $\approx$ `1,179` mol

Đổi `200ml = 0,2l`

`C_{M_{K_3PO_4}}` = `{1,179}/{0,2}` = `5,895M`

b)`n_{CH_4}` = `{12,06 . 10^{22}}/{6 . 10^{23}}` = `0,201` mol

Đổi `800ml = 0,8l`

`C_{M_{CH_4}}` = `{0,201}/{0,8}` = `0,25125M`

c)`n_{Fe_2(SO_4)_3}` = `{250 . 6}/{100 . 400}` = `0,0375` mol

Đổi `400ml = 0,4l`

`C_{M_{Fe_2(SO_4)_3}` = `{0,0375}/{0,4}`= `0,09375M`

2)

a)Đổi `350ml = 0,35l`

`n_{ZnCl_2}` = `0,35 . 2 = 0,7` mol

`C%_{ZnCl_2}` = `{0,7 . 136}/{600}` . `100%` $\approx$ `15,867%`

b)`n_{Ba(OH)_2}` = `4 . 0,3 = 1,2` mol

`C%_{Ba(OH)_2}` = `{1,2 . 171}/{250}` . `100%` =`82,08%`

3)Đổi `200ml = 0,02l`

`n_{Ca_3(PO_4)_2}` = `0,02 . 3 = 0,06` mol

`m_{Ca_3(PO_4)_2}` = `0,06 . 310 = 18,6g`

b)`n_{Al_2(SO_4)_3}` = `4 . 5 = 20` mol

`m_{Al_2(SO_4)_3}` = `20 . 342 = 6840g`