\(n_{KOH}=\dfrac{400.7\%}{56}=0,5\left(mol\right)\)

\(n_{H_2SO_4}=\dfrac{100.19,6\%}{98}=0,2\left(mol\right)\)

PTHH: 2KOH + H₂SO₄ --> K₂SO₄ + 2H₂O

Xét tỉ lệ: \(\dfrac{0,5}{2}>\dfrac{0,2}{1}\) => KOH dư, H₂SO₄ hết

PTHH: 2KOH + H₂SO₄ --> K₂SO₄ + 2H₂O

               0,4<----0,2-------->0,2

=> \(\left\{{}\begin{matrix}m_{KOH\left(dư\right)}=\left(0,5-0,4\right).56=5,6\left(g\right)\\m_{K_2SO_4}=0,2.174=34,8\left(g\right)\end{matrix}\right.\)

m_{dd sau pư} = 400 + 100 = 500 (g)

=> \(\left\{{}\begin{matrix}C\%_{KOH.dư}=\dfrac{5,6}{500}.100\%=1,12\%\\C\%_{K_2SO_4}=\dfrac{34,8}{500}.100\%=6,96\%\end{matrix}\right.\)

\(n_{KOH}=\dfrac{400.7}{100}:56=0,5\left(mol\right)\)

\(n_{H_2SO_4}=\dfrac{100.19,6}{100}:98=0,2\left(mol\right)\)

\(2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O\)

0,4             0,2            0,2

Lập tỉ lệ:

\(\dfrac{0,5}{2}>\dfrac{0,2}{1}\) => KOH dư.

\(m_{dd}=400+100=500\left(g\right)\)

\(n_{KOH.dư}=0,5-0,4=0,1\left(mol\right)\)

\(C\%_{K_2SO_4}=\dfrac{0,2.174.100}{500}=6,96\%\)

\(C\%_{KOH}=\dfrac{0,1.56.100}{500}=1,12\%\)

\(a.Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\\ b.n_{Fe_2O_3}=\dfrac{24}{160}=0,15\left(mol\right)\\ n_{H_2SO_4}=3.0,15=0,45\left(mol\right)\\ m_{ddH_2SO_4}=\dfrac{0,45.98.100}{14,7}=300\left(g\right)\\ m_{ddsau}=24+300=324\left(g\right)\\ n_{Fe_2\left(SO_4\right)_3}=n_{Fe_2O_3}=0,15\left(mol\right)\\ C\%_{ddFe_2\left(SO_4\right)_3}=\dfrac{0,15.400}{324}.100\approx18,519\%\)

\(n_{ZnO}=\dfrac{8,1}{81}=0,1\left(mol\right)\\ n_{H_2SO_4}=\dfrac{19,6\%.200}{98}=0,4\left(mol\right)\\a, ZnO+H_2SO_4\rightarrow ZnSO_4+H_2O\\ b,Vì:\dfrac{0,1}{1}< \dfrac{0,4}{1}\\ \Rightarrow H_2SO_4dư\\ n_{H_2SO_4\left(p.ứ\right)}=n_{ZnSO_4}=n_{ZnO}=0,1\left(mol\right)\\ n_{H_2SO_4\left(dư\right)}=0,4-0,1=0,3\left(mol\right)\\ \Rightarrow m_{H_2SO_4\left(dư\right)}=98.0,3=29,4\left(g\right)\\ c,n_{ZnSO_4}=0,1.161=16,1\left(g\right)\\ m_{ddsau}=m_{ZnO}+m_{ddH_2SO_4}=8,1+200=208,1\left(g\right)\\ \Rightarrow C\%_{ddH_2SO_4\left(dư\right)}=\dfrac{29,4}{208,1}.100\approx14,128\%\\ C\%_{ddZnSO_4}=\dfrac{16,1}{208,1}.100\approx7,737\%\)

ZnO+H₂SO₄->ZnSO₄+H₂O

0,1-----0,1-------0,1-------0,1 mol

n _ZnO=\(\dfrac{8,1}{81}\)=0,1 mol

m _H2SO4 =39,2g =>n _H2SO4=\(\dfrac{39,2}{98}\)=0,4 mol

=>H2SO4 , dư 0,3 mol

=>m _H2SO4=0,3.98=29,4g

=>C%_H2SO4 dư=\(\dfrac{29,4}{200+0,1.18}\).100=14,568%

=>C% _ZnSO4=\(\dfrac{0,1.161}{200+0,1.18}.100=7,9781\%\)

a) Gọi CTHH của ôxit sắt là: Fe_xO_y

Theo đề bài ra ta có: 56x : (56x + 16y) = 70%

<=> 5600x = 3920x + 1120y

<=> 1680x = 1120y => x:y = 2 : 3

Công thức hoá học của ôxit sắt là: Fe₂O₃

b)  PTHH:   Fe₂O₃ + 3H₂SO₄ → Fe₂(SO₄)₃ + 3H₂ ↑

Khối lượng chất tan H₂SO₄ là: 200 . 19,6% = 39,2 gam

Số mol của H₂SO₄ là: 39,2 : 98 = 0,4 mol

Số mol của Fe₂O₃ là: 2/15 mol => m_Fe2O3 = 21,3 gam

Số mol của Fe₂(SO₄)₃ là: 2/15 => m_Fe2(so4)3 = 53,3 gam

Số mol của H₂ là: 0,4 => m_H2 = 0,8 gam

Khối lượng dung dịch sau phản ứng là: 

 21,3 + 200 - 0,8 = 220,5 gam

C% = (53,3 : 220,5).100% = 24,2% 

\(a,n_{Zn}=\dfrac{19,5}{65}=0,3\left(mol\right)\\ m_{H_2SO_4}=200.19,6\%=39,2\left(g\right)\\ \rightarrow n_{H_2SO_4}=\dfrac{39,2}{98}=0,4\left(mol\right)\)

PTHH: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\uparrow\)

bđ        0,3    0,4

pư        0,3        0,3

spư      0           0,1           0,3        0,3

\(\rightarrow V_{H_2}=0,3.22,4=6,72\left(l\right)\)

\(b,m_{dd}=19,5+200-0,3.2=218,9\left(g\right)\\ \rightarrow\left\{{}\begin{matrix}C\%_{ZnSO_4}=\dfrac{0,3.161}{218,9}.100\%=22,06\%\\C\%_{H_2SO_4\left(dư\right)}=\dfrac{0,1.98}{218,9}.100\%=4,48\%\end{matrix}\right.\)

Tham Khảo

\(m_{ct}=\dfrac{19,6.200}{100}=39,2\left(g\right)\)

\(n_{H2SO4}=\dfrac{39,2}{98}=0,4\left(mol\right)\)

\(m_{ct}=\dfrac{5,2.200}{100}=10,4\left(g\right)\)

\(n_{BaCl2}=\dfrac{10,4}{208}=0,05\left(mol\right)\)

Pt : \(H_2SO_4+BaCl_2\rightarrow2HCl+BaSO_4|\)

            1             1             2             1

          0,4           0,05        0,1         0,05

a) Lập tỉ số so sánh: \(\dfrac{0,4}{1}>\dfrac{0,05}{1}\)

             ⇒ H₂SO₄ dư , BaCl₂ phản ứng hết

             ⇒ Tính toán dựa vào số mol của BaCl₂

\(n_{BaSO4}=\dfrac{0,05.1}{1}=0,05\left(mol\right)\)

⇒ \(m_{BaSO4}=0,05.233=11,65\left(g\right)\)

b) \(n_{HCl}=\dfrac{0,05.2}{1}=0,1\left(mol\right)\)

⇒ \(m_{HCl}=0,1.36,5=3,65\left(g\right)\)

\(n_{H2SO4\left(dư\right)}=0,4-0,05=0,35\left(mol\right)\)

⇒ \(m_{H2SO4\left(dư\right)}=0,35.98=34,3\left(g\right)\)

\(m_{ddspu}=200+200-11,65=388,35\left(g\right)\)

\(C_{ddHCl}=\dfrac{3,65.100}{388,35}=0,94\)⁰/₀

\(C_{ddH2SO4\left(dư\right)}=\dfrac{34,3.100}{388,35}=8,83\)⁰/₀

 Chúc bạn học tốt

\(a.n_{H_2SO_4}=\dfrac{200.19,6\%}{98}=0,4\left(mol\right)\\ n_{BaCl_2}=\dfrac{200.5,2\%}{208}=0,05\left(mol\right)\\ BaCl_2+H_2SO_4\rightarrow BaSO_4\downarrow+2HCl\\ Vì:\dfrac{0,05}{1}< \dfrac{0,4}{1}\\ \Rightarrow H_2SO_4dư\\ n_{BaSO_4}=n_{BaCl_2}=0,05\left(mol\right)\\ \Rightarrow m_{\downarrow}=m_{BaSO_4}=0,05.233=11,65\left(g\right)\\ b.m_{ddsau}=200+200-11,65=388,35\left(g\right)\\ C\%_{ddHCl}=\dfrac{0,05.2.36,5}{388,35}.100\approx0,94\%\\ C\%_{ddH_2SO_4\left(dư\right)}=\dfrac{\left(0,4-0,05\right).98}{388,35}.100\approx8,832\%\)

\(n_{H_2SO_4}=\dfrac{200.19,6}{100.98}=0,4mol\\ CuO+H_2SO_4\rightarrow CuSO_4+H_2O\\ n_{CuSO_4\left(A\right)}=n_{CuO}=n_{H_2SO_4}=0,4mol\\ n_{Cu\left(OH\right)_2}=\dfrac{29,4}{98}=0,3mol\\ CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2+Na_2SO_4\\\Rightarrow\dfrac{0,4}{1}>\dfrac{0,3}{1}\Rightarrow CuSO_4.pư.không.hết\)

\(CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2+Na_2SO_4\)

0,3mol        0,6mol                                        0,3mol

\(m_{ddB}=0,4.80+200+0,6.40-29,4=226,6g\\ C_{\%Na_2SO_4\left(B\right)}=\dfrac{0,3.142}{226,6}\cdot100=18,8\%\)