a, \(n_{NaOH}=0,2.1=0,2\left(mol\right)\)

\(m_{NaOH}=0,2.40=8\left(g\right)\)

b, \(n_{H_2SO_4}=2.0,1=0,2\left(mol\right)\)

\(c,C\%=\dfrac{6}{200}.100\%=3\%\)

\(m_{NaCl}=\dfrac{200.8}{100}=16\left(g\right)\)

a) m_H2SO4=​​490x20%= 98g      → n_H2SO4= 1 mol

    m_BaCl2=41,6g                      → n_BaCl2=0,2 mol

                       BaCl₂      +     H₂SO₄         →      BaSO₄↓       +       2HCl

ban đầu:       0,2 mol             1 mol  

PƯ:               0,2 mol             0,2 mol                0,2 mol                  0,4 mol

Còn lại            0 mol              0,8 mol                0,2 mol                   0,4 mol

Như vậy: kết tủa A: 0,2 mol BaSO₄; Dung dịch B: 0,8 mol H₂SO₄ dư và 0,4 mol HCl

→m_A= m_BaSO4= 0,2 x 233 = 46,6g

m_{dung dịch B}= m_{dung dịch H2SO4} + m_{dung dịch BaCl2} - m_↓BaSO4

                  = 490 + 800 - 46,6 = 1243,4g

%H₂SO₄ = ( 0,8 x 98)/1243,4 x100% = 6,31%

%HCl = ( 0,4 x 36,5)/1243,4 x100% = 1,17%

b)Trong 1000g dung dịch NaCl 15% có: m_NaCl= 1000 x 15% = 150g

→m_H2O= m_{dung môi} = m_{dung dịch} - m_{chất tan} = 1000 - 150 = 850g

Câu 1 :

mNaCl = 5.85*100/100= 5.85 g

mH2SO4 = 100*49/100=49 g

mNaOH = 100*20/100=20 g

BaSO4 không tan trong nước nên không tính được

Câu 2 :

nNa = 2.3/23 = 0.1 mol

Na + H2O --> NaOH + 1/2H2

0.1___________0.1______0.05

mdd sau phản ứng = 2.3 + 100 - 0.1 = 102.2 g

mNaOH = 4g

C%NaOH = 4/102.2*100% = 3.91%

Câu 3 :

nCa = 0.05 mol

Ca + 2H2O --> Ca(OH)2 + 2H2

0.05____________0.05_____0.1

mdd = 100 + 2 - 0.2 = 101.8 g

mCa(OH)2 = 3.7 g

C%Ca(OH)2 = 3.63%

Câu 4 :

nH2 = 5*10^-5 mol

Ca + 2H2O --> Ca(OH)2 + H2

5*10^-5__________5*10^-5__5*10^-5

mCa = 0.002 g

mCa(OH)2 = 0.0037 g

mdd sau phản ứng = 100 g

C%Ca(OH)2 = 0.0037%

a.

 \(m_{Ag}=m_{k.tan}=8,7\left(g\right)\\ m_{Zn,Mg}=20-8,7=11,3\left(g\right)\\ \left\{{}\begin{matrix}n_{Zn}=a\left(mol\right)\\n_{Mg}=b\left(mol\right)\end{matrix}\right.\left(a,b>0\right)\\Zn+H_2SO_4\rightarrow ZnSO_4+H_2\\ Mg+H_2SO_4\rightarrow MgSO_4+H_2\\ \rightarrow\left\{{}\begin{matrix}65a+24b=11,3\\a+b=0,3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,2\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\%m_{Ag}=\dfrac{8,7}{20}.100=43,5\%\\\%m_{Mg}=\dfrac{24.0,2}{20}.100=24\%\\\%m_{Zn}=\dfrac{0,1.65}{20}.100=32,5\%\end{matrix}\right.\)

b. 

\(n_{H_2SO_4\left(tổng\right)}=a+b=0,3\left(mol\right)\\ V_{ddH_2SO_4\left(tổng\right)}=\dfrac{0,3}{0,5}=0,6\left(lít\right)=600\left(ml\right)\)

PTHH: \(Na_2CO_3+2HCl\rightarrow2NaCl+CO_2+H_2O\\ 0,1mol:0,2mol\rightarrow0,2mol:0,1mol:0,1mol\)

\(n_{Na_2CO_3}=\dfrac{10,6}{106}=0,1\left(mol\right)\)

a. \(n_{NaCl}=0,2.58,5=11,7\left(g\right)\)

b. \(V_{CO_2}=0,1.22,4=2,24\left(l\right)\)

c. \(m_{HCl}=0,2.36,5=7,3\left(g\right)\)

\(n_{Na_2CO_3}=\dfrac{m_{Na_2CO_3}}{M_{Na_2CO_3}}=\dfrac{10,6}{106}=0,1\left(mol\right)\)

PTHH:

\(Na_2CO_3+2HCl\rightarrow2NaCl+H_2O+CO_2\)

\(0,1............0,2........0,2.......0,1.......0,1\) (mol)

a. \(m_{NaCl}=M_{NaCl}.n_{NaCl}=0,2.58,5=11,7\left(g\right)\)

b. \(V_{CO_2}=22,4.n_{CO_2}=22,4.0,1=2,24\left(l\right)\)

c. \(m_{HCl}=n_{HCl}.M_{HCl}=0,2.36,5=7,3\left(g\right)\)

\(n_{Na_2CO_3}=\dfrac{10,6}{106}=0,1\left(mol\right)\)
PTHH: \(Na_2CO_3+2HCl\rightarrow2NaCl+H_2O+CO_2\)

a. Theo PT ta có: \(n_{NaCl}=\dfrac{0,1.2}{1}=0,2\left(mol\right)\)
\(\Rightarrow m_{NaCl}=0,2.58,5=11,7\left(g\right)\)
b. Theo PT ta có: \(n_{CO_2}=n_{Na_2CO_3}=0,1\left(mol\right)\)
\(\Rightarrow V_{CO_2}=0,1.22,4=2,24\left(l\right)\)
c. Theo PT ta có: \(n_{HCl}=\dfrac{0,1.2}{1}=0,2\left(mol\right)\)
\(\Rightarrow m_{HCl}=0,2.36,5=7,3\left(g\right)\)

\(M_{KNO_3}=39+14+16\cdot3=101g/mol\\ M_{H_2SO_4}=1\cdot2+32+16\cdot4=98g/mol\\ M_{CuSO_4}=64+32+16\cdot4=160g/mol\\ M_{Al\left(OH\right)_3}=27+\left(1+17\right)\cdot3=78g/mol\\ M_{NaCl}=23+35,5=58,5g/mol\)

\(M_{KNO_3}=39+14+16\cdot3=101g/mol\\ M_{H_2SO_4}=1\cdot2+32+16\cdot4=98g/mol\\ M_{CuSO_4}=64+32+16\cdot4=160g/mol\\ M_{Al\left(OH\right)}_{_3}=27+\left(1+16\right)\cdot3=78g/mol\\ M_{NaCl}=23+35,5=58,5g/mol\)

khí X có lẽ là H2 và SO2