a, \(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)

b, \(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\)

\(n_{H_2SO_4}=\dfrac{200.8\%}{98}=\dfrac{8}{49}\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{\dfrac{8}{49}}{1}\), ta được H₂SO₄ dư.

Theo PT: \(n_{H_2SO_4\left(pư\right)}=n_{MgSO_4}=n_{H_2}=n_{Mg}=0,1\left(mol\right)\)

\(\Rightarrow m_{H_2SO_4\left(dư\right)}=\left(\dfrac{8}{49}-0,1\right).98=6,2\left(g\right)\)

c, \(C\%_{MgSO_4}=\dfrac{0,1.120}{2,4+200-0,1.2}.100\%\approx5,93\%\)

a) \(n_{CO_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

PTHH: CO₂ + 2NaOH → Na₂CO₃ + H₂O

Mol:      0,15       0,3              0,15

\(C_{M_{ddNaOH}}=\dfrac{0,3}{0,2}=1,5M\)

b) Na₂CO₃: natri cacbonat

\(m_{Na_2CO_3}=0,15.106=15,9\left(g\right)\)

c) 

PTHH: 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O

Mol:       0,15          0,075

\(V_{ddH_2SO_4}=\dfrac{0,075}{1}=0,075\left(l\right)=75\left(ml\right)\)

\(n_{CO2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

a) Pt : \(CO_2+2NaOH\rightarrow Na_2CO_3+H_2O|\)

             1            2                   1               1

           0,15       0,3                0,15

\(n_{NaOH}=\dfrac{0,15.2}{1}=0,3\left(mol\right)\)

200ml = 0,2l

\(C_{M_{ddNaOH}}=\dfrac{0,3}{0,2}=1,5\left(M\right)\)

b \(n_{Na2CO3}=\dfrac{0,3.1}{2}=0,15\left(mol\right)\)

⇒ \(m_{Na2CO3}=0,15.106=15,9\left(g\right)\)

Ten muoi thu duoc : natri cacbonat 

c) Pt : \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O|\)

                 2               1               1                2

              0,3              0,15

\(n_{H2SO4}=\dfrac{0,3.1}{2}=0,15\left(mol\right)\)

\(V_{ddH2SO4}=\dfrac{0,15}{1}=0,15\left(l\right)=150\left(ml\right)\)

 Chuc ban hoc tot

\(n_{CuO}=\dfrac{1,6}{80}=0,02\left(mol\right)\)

\(m_{H_2SO_4}=100.20\%=20\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{20}{98}=\dfrac{10}{49}\left(mol\right)\)

PTHH: CuO + H₂SO₄ → CuSO₄ + H₂O

Mol:      0,02      0,02          0,02

Ta có: \(\dfrac{0,02}{1}< \dfrac{\dfrac{10}{49}}{1}\) ⇒ CuO hết, H₂SO₄ dư

m_{dd sau pứ} = 1,6 + 100 = 101,6 (g)

\(C\%_{ddCuSO_4}=\dfrac{0,02.160.100\%}{101,6}=3,15\%\)

\(C\%_{ddH_2SO_4}=\dfrac{\left(\dfrac{10}{49}-0,02\right).98.100\%}{101,6}=17,76\%\)

PTHH: \(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)

Ta có: \(n_{Fe_2O_3}=\dfrac{4,8}{160}=0,03\left(mol\right)\) 

\(\Rightarrow\left\{{}\begin{matrix}n_{H_2SO_4}=0,09\left(mol\right)\\n_{Fe_2\left(SO_4\right)_3}=0,03\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{ddH_2SO_4}=\dfrac{0,09\cdot98}{9,8\%}=90\left(g\right)\\m_{Fe_2\left(SO_4\right)_3}=0,03\cdot400=12\left(g\right)\end{matrix}\right.\)

\(a.HCl+NaOH\rightarrow NaCl+H_2O\)

PỨ trung hoà 

\(b,n_{NaOH}=0,1.1=0,1mol\\ n_{NaCl}=n_{NaOH}=n_{HCl}0,1mol\\ m=m_{HCl}=0,1.36,5=3,65g\\ c,m_{NaCl}=0,1.58,5=5,85g\\ d,n_{HCl}=\dfrac{73.10}{100.36,5}=0,2mol\\ \Rightarrow\dfrac{0,1}{1}< \dfrac{0,2}{1}\Rightarrow HCl.dư\\ n_{HCl,pứ}=n_{NaOH}=0,1mol\\ m_{HCl,dư}=\left(0,2-0,1\right).36,5=3,65g\)

\(n_{MgCO_3}=\dfrac{16,8}{84}=0,2mol\\ a.MgCO_3+H_2SO_4->MgSO_4+H_2O+CO_2\\ 2NaOH+H_2SO_{\text{4 }}->Na_2SO_4+2H_2O\\ b.n_{H_2SO_4dư}=\dfrac{1}{2}n_{NaOH}=\dfrac{1}{2}.80.0,1:40=0,1mol\\ n_{H_2SO_4\left(MgCO_3\right)}=0,2mol\\ c.C\%=\dfrac{98.0,3}{200}.100\%=14,7\%\\ V=0,2.22,4=4,48L\\ d.m_{ddsau}=200+16,8-44.0,2+80=288g\\ C\%_{Na_2SO_4}=\dfrac{40.0,1}{288}.100\%=1,39\%\\ C\%_{MgSO_4}=\dfrac{120.0,2}{288}.100\%=8,33\%\)

\(n_{MgCO_3}=\dfrac{16,8}{84}=0,2\left(mol\right)\)

\(n_{NaOH}=\dfrac{80}{40}=2\left(mol\right)\)

PTHH :

\(MgCO_3+H_2SO_4\rightarrow MgSO_4+H_2O+CO_2\uparrow\)

0,2              0,2                                         0,2 

\(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)

2                 1                 1

Vậy có 0,2 mol H2SO4 phản ứng với MgCO3 

       có 1 mol H2SO4 phản ứng với NaOH 

\(m_{H_2SO_4}=1,2.98=117,6\left(g\right)\)

\(c,C\%_{H_2SO_4}=\dfrac{117,6}{200}.100\%=58,8\%\)

\(V_{CO_2}=0,2.22,4=4,48\left(l\right)\)

\(d,m_{Na_2SO_4}=1.142=142\left(g\right)\)

\(m_{ddNaOH}=\dfrac{80.100}{10}=800\left(g\right)\)

\(m_{ddH_2SO_4dư}=1.98:58,8\%\approx166,67\left(g\right)\)

\(m_{ddNa_2SO_4}=800+166,67=966,67\left(g\right)\)

\(C\%_{Na_2SO_4}=\dfrac{142}{966,67}.100\%\approx14,69\%\)

a) PTHH: \(FeCl_3+3NaOH\rightarrow3NaCl+Fe\left(OH\right)_3\downarrow\)

                 \(2Fe\left(OH\right)_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+6H_2O\)

b) Ta có: \(n_{FeCl_3}=0,3\cdot0,5=0,15\left(mol\right)\)

\(\Rightarrow n_{NaOH}=0,45mol\) \(\Rightarrow V_{ddNaOH}=\dfrac{0,45}{0,25}=1,8\left(l\right)\)

c) Theo PTHH: \(n_{NaCl}=n_{NaOH}=0,45mol\)

\(\Rightarrow C_{M_{NaCl}}=\dfrac{0,45}{2,1}\approx0,21\left(M\right)\) 

(Coi như thể tích dd thay đổi không đáng kể)

d) Theo PTHH: \(n_{H_2SO_4}=\dfrac{3}{2}n_{Fe\left(OH\right)_3}=\dfrac{3}{2}n_{FeCl_3}=0,225mol\)

\(\Rightarrow m_{ddH_2SO_4}=\dfrac{0,225\cdot98}{20\%}=110,25\left(g\right)\) 

\(\Rightarrow V_{ddH_2SO_4}=\dfrac{110,25}{1,14}\approx96,71\left(ml\right)\)