a)

R+H_2SO_4->RSO_4+H_2

Theo PT: n_R=n_{H_2}={2,8}/{22,4}=0,125(mol)

->M_R=3/{0,125}=24(g//mol)

->R: magnesium (Mg).

b)

Mg+H_2SO_4->MgSO_4+H_2

Theo PT: n_{MgSO_4}=n_{H_2SO_4\ pu}=n_{H_2}=0,125(mol)

n_{NaOH}={50.1,25.20\%}/{40}=0,3125(mol)

H_2SO_4+2NaOH->Na_2SO_4+2H_2O

MgSO_4+2NaOH->Mg(OH)_2+Na_2SO_4

Theo PT: n_{NaOH}=2n_{MgSO_4}+2n_{H_2SO_4\ du}

->n_{H_2SO_4\ du}={0,3125-0,125.2}/2=0,03125(mol)

->\sum n_{H_2SO_4}=0,125+0,03125=0,15625(mol)

->x={0,15625.98}/{200}.100\%=7,65625(\%)

c)

n_{Na_2SO_4}=1/2n_{NaOH}=0,15625(mol)

n_{Mg(OH)_2}=n_{MgSO_4}=0,125(mol)

m_{dd\ A}=3+200-0,125.2=202,75(g)

->C\%_{MgSO_4}={0,125.120}/{202,75}.100\%\approx 7,4\%

        C\%_{H_2SO_4\ du}={0,03125.98}/{202,75}.100\%\approx 1,51\%

m_{dd\ B}=200+3+50.1,25-0,125.2-0,125.58=258(g)

->C\%_{Na_2SO_4}={0,15625.142}/{258}.100\%\approx 8,6\%