Câu 1 . 

`n_(S) = 0.64 / 32 = 0.02 ( mol ) `

Bảo toàn mol e :

   `   2n_(Zn) = 6n_(S)`

 `->   n_(Zn) = 3n_S = 3 × 0.02 = 0.06 ( mol ) `

`=> m_(Zn) = 0.06 × 65 = 3.9 ( g ) `

`-> m_(ZnSO_4) = 0.06 × 161 = 9.66 ( g ) `

Câu 2 . 

`n_(SO_2) = 6.72 / 22.4 = 0.3 ( mol ) `

`n_(S) = 1.6 / 32 = 0.05 ( mol ) `

Bảo toàn e : 

   ` 2n_(Mg) = 2n_(SO_2)  +  6n_(S)`

`-> n_(Mg) = n_(SO_2)   +  3n_(S) `

`-> n_(Mg)  =   0.3   + 3 × 0.05 = 0.45 ( mol ) `

`m_(Mg) = 0.45 × 24 = 10.8 ( g ) `

1)

Rắn màu vàng là \(S\)

\({n_S} = \frac{{0,64}}{{32}} = 0,02{\text{ mol}}\)

 Bảo toàn e:

\(2{n_{Zn}} = 6{n_S} \to {n_{Zn}} = 3{n_S} = 0,02.3 = 0,06{\text{ mol = }}{{\text{n}}_{ZnS{O_4}}}\)

\( \to m = {m_{Zn}} = 0,06.65 = 3,9{\text{ }}gam\)

\({m_{ZnS{O_4}}} = 0,06.(65 + 96) = 9,66{\text{ gam}}\)

2)

Ta có:

\({n_{S{O_2}}} = \frac{{6,72}}{{22,4}} = 0,3{\text{ mol;}}{{\text{n}}_S} = \frac{{1,6}}{{32}} = 0,05{\text{ mol}}\)

Bảo toàn e:

\(2{n_{Mg}} = 2{n_{S{O_2}}} + 6{n_S} = 0,3.2 + 0,05.6 = 0,9{\text{ mol}}\)

\( \to {n_{Mg}} = 0,45{\text{ mol}} \to {\text{m = }}{{\text{m}}_{Mg}} = 0,45.24 = 10,8{\text{ gam}}\)