a. PTHH: H₂SO₄ + 2NaOH ---> Na₂SO₄ + 2H₂O

b. Ta có: \(C_{\%_{H_2SO_4}}=\dfrac{m_{H_2SO_4}}{300}.100\%=19,6\%\)

=> \(m_{H_2SO_4}=58,8\left(g\right)\)
=> \(n_{H_2SO_4}=\dfrac{58,8}{98}=0,6\left(mol\right)\)

Ta lại có: \(C_{\%_{NaOH}}=\dfrac{m_{NaOH}}{200}.100\%=20\%\)

=> m_NaOH = 40(g)

=> \(n_{NaOH}=\dfrac{40}{40}=1\left(mol\right)\)

Ta thấy: \(\dfrac{0,6}{1}>\dfrac{1}{2}\)

Vậy H₂SO₄ dư.

=> \(m_{dd_{Na_2SO_4}}=300+40=340\left(g\right)\)

Theo PT: \(n_{Na_2SO_4}=\dfrac{1}{2}.n_{NaOH}=\dfrac{1}{2}.1=0,5\left(mol\right)\)

=> \(m_{Na_2SO_4}=0,5.142=71\left(g\right)\)

=> \(C_{\%_{Na_2SO_4}}=\dfrac{71}{340}.100\%=20,88\%\)

mBaCl2=10.4(g)

nBaCl2=0.05(mol)

mH2SO4=11.76(g)

nH2SO4=0.12(mol)

BaCl2+H2SO4->BaSO4+2HCl

Theo pthh:nH2SO4=nBaCl2

theo bài ra,nH2SO4>nBaCl2

->H2SO4 dư

nH2SO4 dư=0.12-0.05=0.07(mol)

mH2SO4 dư=0.07*98=6.86(g)

nBaSO4=0.05(mol)

mBaSO4=11.65(g)

nHCl=0.05*2=0.1(mol)

mHCl=3.65(g)

mdd sau phản ứng:200+58.8-11.65=247.15(g)

C%(HCl)=3.65:247.15*100=1.48%

C%(H2SO4)=6.86:247.15*100=2.78%

đề bài có vấn đề=.=

À, là 100 g dung dịch H2SO4 

biết ngay mà

\(n_{Zn}=\dfrac{1,95}{65}=0,03\left(mol\right)\\ m_{H_2SO_4}=\dfrac{22,05.20}{100}=4,41\left(g\right)\\ n_{H_2SO_4}=\dfrac{4,41}{98}=0,045\left(mol\right)\\ pthh:Zn+H_2SO_4\rightarrow ZnSO_4+H_2\uparrow\) 
\(LTL:\dfrac{0,03}{1}< \dfrac{0,045}{1}\) 
=> H₂SO₄ dư 
\(n_{H_2SO_4\left(p\text{ư}\right)}=n_{ZnSO_4}=n_{H_2}=n_{Zn}=0,03\left(mol\right)\\ m_{H_2SO_4\left(d\right)}=\left(0,045-0,03\right).98=1,47\left(g\right)\\ m_{\text{dd}}=1,95+22,05-\left(0,03.2\right)=23,94\left(g\right)\\ C\%_{ZnCl_2}=\dfrac{0,03.136}{23,94}.100\%=17\%\)

\(a,n_{Zn}=\dfrac{1,95}{65}=0,03\left(mol\right)\\ n_{H_2SO_4}=\dfrac{22,05}{98}=0,225\left(mol\right)\)

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

bđ       0,03   0,225

pư       0,03    0,03 

spư     0         0,195       0,03        0,03

\(b,m_{H_2SO_4\left(dư\right)}=0,195.98=19,11\left(g\right)\\ c,m_{dd}=1,95+22,05-0,03.2=23,94\left(g\right)\\ C\%_{ZnSO_4}=\dfrac{0,03.161}{23,94}.100\%=20,18\%\)

Bài 1:

a, Hiện tượng: Có khí mùi hắc thoát ra.

b, Ta có: \(m_{H_2SO_4}=100.24,5\%=24,5\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{24,5}{98}=0,25\left(mol\right)\)

PT: \(Na_2SO_3+H_2SO_4\rightarrow Na_2SO_4+SO_2+H_2O\)

Theo PT: \(n_{Na_2SO_3}=n_{H_2SO_4}=0,25\left(mol\right)\)

\(\Rightarrow C\%_{Na_2SO_3}=\dfrac{0,25.126}{200}.100\%=15,75\%\)

c, Theo PT: \(n_{SO_2}=n_{Na_2SO_4}=n_{H_2SO_4}=0,25\left(mol\right)\)

⇒ m _{dd sau pư} = 200 + 100 - 0,25.64 = 284 (g)

\(\Rightarrow C\%_{Na_2SO_4}=\dfrac{0,25.142}{284}.100\%=12,5\%\)

Bài 2:

a, Hiện tượng: Xuất hiện kết tủa trắng.

PT: \(BaCl_2+MgSO_4\rightarrow MgCl_2+BaSO_{4\downarrow}\)

b, Ta có: \(m_{BaCl_2}=200.20,8\%=41,6\left(g\right)\Rightarrow n_{BaCl_2}=\dfrac{41,6}{208}=0,2\left(mol\right)\)

Theo PT: \(n_{MgCl_2}=n_{BaSO_4}=n_{MgSO_4}=n_{BaCl_2}=0,2\left(mol\right)\)

\(\Rightarrow m_{ddMgSO_4}=\dfrac{0,2.120}{12\%}=200\left(g\right)\)

c, Ta có: m _{dd sau pư} = 200 + 200 - 0,2.233 = 353,4 (g)

\(\Rightarrow C\%_{MgCl_2}=\dfrac{0,2.95}{353,4}.100\%\approx5,38\%\)

\(n_{Fe_2O_3}=\dfrac{20}{160}=0,125\left(mol\right)\)

PTHH:

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

0,125     0,375             0,125           0,375 

\(m_{ddH_2SO_4}=\dfrac{0,375.98.100}{25}=147\left(g\right)\)

\(C\%_{Fe_2\left(SO_4\right)_3}=\dfrac{0,125.406}{20+147}\approx30,39\%\)

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

\(m_{NaOH}= 20 .10\)%=2g \(\Rightarrow n_{NaOH}=\dfrac{2}{40}=0,05 mol\)

m_H2SO4= 20 . 10% = 2g \(\Rightarrow n_{H_2SO_4}= \dfrac{2}{98}= 0,02 mol\)

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

Trước pư:  0,05            0,02

PƯ:            0,04            0,02          0,02

Sau pư:      0,01              0             0,02

dd sau pư gồm NaOH dư và Na₂SO₄

\(m_{dd sau pư}= m_{NaOH} + m_{H_2SO_4}= 20 + 20=40g\)

Ta có

C%_\(NaOH\)=\(\dfrac{0,01 . 40}{40} . 100\)%=1%

C%_{\(Na_2SO_4\)}=\(\dfrac{0,02 .142}{40} . 100\)%=7,1%

Bài 2:

a) PTHH: \(Na_2O+H_2O\rightarrow2NaOH\)

b) Dung dịch A là dung dịch bazơ

Ta có: \(n_{Na_2O}=\dfrac{3,1}{62}=0,05\left(mol\right)\) \(\Rightarrow n_{NaOH}=0,1\left(mol\right)\) \(\Rightarrow C_{M_{NaOH}}=\dfrac{0,1}{1}=0,1\left(M\right)\)

c) Sửa đề: dd H₂SO₄ 9,8%

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

Theo PTHH: \(n_{H_2SO_4}=\dfrac{1}{2}n_{NaOH}=0,05\left(mol\right)\)

\(\Rightarrow m_{ddH_2SO_4}=\dfrac{0,05\cdot98}{9,8\%}=50\left(g\right)\) \(\Rightarrow V_{ddH_2SO_4}=\dfrac{50}{1,14}\approx43,86\left(ml\right)\)

Bài 1:

PTHH: \(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)

Ta có: \(\left\{{}\begin{matrix}n_{CuO}=\dfrac{16}{80}=0,2\left(mol\right)\\n_{H_2SO_4}=\dfrac{200\cdot19,6\%}{98}=0,4\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\) Axit còn dư

\(\Rightarrow n_{CuSO_4}=0,2\left(mol\right)=n_{H_2SO_4\left(dư\right)}\)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{CuSO_4}=\dfrac{0,2\cdot160}{200+16}\cdot100\%\approx14,81\%\\C\%_{H_2SO_4\left(dư\right)}=\dfrac{0,2\cdot98}{200+16}\cdot100\%\approx9,07\%\end{matrix}\right.\)