a, \(Na_2SO_3+H_2SO_4\rightarrow Na_2SO_4+SO_2+H_2O\)

\(K_2SO_3+H_2SO_4\rightarrow K_2SO_4+SO_2+H_2O\)

Ta có: \(n_{SO_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

Theo PT: \(n_{H_2SO_4}=n_{SO_2}=0,3\left(mol\right)\)

\(\Rightarrow m_{ddH_2SO_4}=\dfrac{0,3.98}{20\%}=147\left(g\right)\)

b, Ta có: 126n_Na2SO3 + 158n_K2SO3 = 44,2 (1)

Theo PT: \(n_{SO_2}=n_{Na_2SO_3}+n_{K_2SO_3}=0,3\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{Na_2SO_3}=0,1\left(mol\right)\\n_{K_2SO_3}=0,2\left(mol\right)\end{matrix}\right.\)

Có: m _{dd sau pư} = 44,2 + 147 - 0,3.64 = 172 (g)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{Na_2SO_3}=\dfrac{0,1.126}{172}.100\%\approx7,33\%\\C\%_{K_2SO_3}=\dfrac{0,2.158}{172}.100\%\approx18,37\%\end{matrix}\right.\)

c, \(n_{Ba\left(OH\right)_2}=0,5.1=0,5\left(mol\right)\)

\(\Rightarrow\dfrac{n_{SO_2}}{n_{Ba\left(OH\right)_2}}=0,6< 1\) → Pư tạo BaSO₃.

PT: \(SO_2+Ba\left(OH\right)_2\rightarrow BaSO_3+H_2O\)

\(n_{BaSO_3}=n_{SO_2}=0,3\left(mol\right)\Rightarrow m_{BaSO_3}=0,3.217=65,1\left(g\right)\)

a) Gọi \(\left\{{}\begin{matrix}n_{Zn}=a\left(mol\right)\\n_{Al}=b\left(mol\right)\end{matrix}\right.\left(\text{Đ}K:a,b>0\right)\)

PTHH: Zn + H₂SO₄ ---> ZnSO₄ + H₂ 

           a------>a---------->a----------->a

           2Al + 3H₂SO₄ ---> Al₂(SO₄)₃ + 3H₂ 

           b----->1,5b--------->0,5b------->1,5a

=> \(\left\{{}\begin{matrix}65a+27b=20,3\\161b+0,5a.342=65,9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,25\\b=0,15\end{matrix}\right.\)

=> \(V=V_{H_2}=\left(0,25+0,15.1,5\right).22,4=10,64\left(l\right)\)

b) \(\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{0,25.65}{20,3}.100\%=80,05\%\\\%m_{Al}=100\%-80,05\%=19,95\%\end{matrix}\right.\)

c) \(m_{\text{dd}H_2SO_4}=\dfrac{\left(0,25+1,5.0,15\right).98}{10\%}=465,5\left(g\right)\)

PTHH: \(2Na+H_2SO_4\rightarrow Na_2SO_4+H_2\uparrow\)  (1)

            \(Na_2SO_4+Ba\left(OH\right)_2\rightarrow2NaOH+BaSO_4\downarrow\)  (2)

            \(2NaOH+MgCl_2\rightarrow2NaCl+Mg\left(OH\right)_2\downarrow\)  (3)

            \(Mg\left(OH\right)_2\xrightarrow[]{t^o}MgO+H_2O\)  (4)

Ta có: \(\left\{{}\begin{matrix}n_{Na_2SO_4}=\dfrac{1}{2}n_{Na}=\dfrac{1}{2}\cdot\dfrac{2,3}{23}=0,05\left(mol\right)\\n_{BaCl_2}=\dfrac{60\cdot14,25\%}{208}=0,05\left(mol\right)\\n_{MgCl_2}=\dfrac{30\cdot19\%}{95}=0,06\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\) PT (2) p/ứ hết; PT (3) có MgCl₂ dư 0,01 mol  

\(\Rightarrow n_{MgO}=n_{Mg\left(OH\right)_2}=n_{BaSO_4}=0,05\left(mol\right)\)

\(\Rightarrow m_{rắn}=m_{MgO}+m_{BaSO_4}=0,05\cdot\left(40+233\right)=13,65\left(g\right)\)

b) Theo các PTHH: \(\left\{{}\begin{matrix}n_{NaCl}=n_{Na}=0,1\left(mol\right)\\n_{Mg\left(OH\right)_2}=0,05\left(mol\right)=n_{H_2SO_4}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{NaCl}=0,1\cdot58,5=5,85\left(g\right)\\m_{ddH_2SO_4}=\dfrac{0,05\cdot98}{4,9\%}=100\left(g\right)\\m_{Mg\left(OH\right)_2}=0,05\cdot58=2,9\left(g\right)\end{matrix}\right.\)

Mặt khác: \(m_{dd\left(sau.p/ứ\right)}=m_{Na}+m_{ddH_2SO_4}+m_{ddBaCl_2}+m_{ddMgCl_2}-m_{BaSO_4}-m_{Mg\left(OH\right)_2}=177,75\left(g\right)\)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{NaCl}=\dfrac{5,85}{177,75}\cdot100\%\approx3,29\%\\C\%_{MgCl_2\left(dư\right)}=\dfrac{0,01\cdot95}{177,75}\cdot100\%\approx0,53\%\end{matrix}\right.\) 

Na+ H2O=> NaOH+ 1/2 H2

nNa= 4,6/23=0,2( mol)

nH2=0,2* 1/2=0,1 => V=2,24(l)

b, C%=0,8%

c, hình như thừa hoặc thiếu đầu bài. m=8g

\(n_{Na}=\dfrac{m}{M}=\dfrac{4,6}{23}=0,2\left(mol\right)\)

\(n_{ddH_2SO_4}=\dfrac{m}{M}=\dfrac{200}{98}=2\)

PTHH:\(2Na+H_2SO_4\rightarrow Na_2SO_4+H_2\)

tpư:     0,2          2        

pư:      0,2          0,1           0,1           0,1

spư:      0           1,9            0,1         0,1

a)\(V_{H_2}=n.22,4\)=0,1.22,4=2,24

b)\(m_{Na_2SO_4}=n.M\)=0,1.142=14,2

\(m_{H_2SO_4dư}=n.M\)=1,9.98=186,2

c)\(C\%H_2SO_4=\dfrac{m_{ct}}{m_{dd}}.100=\dfrac{0,1.98}{200}.100\)=0,099%

Đáp án C

n_Cu = 0,13 => Y hòa tan vừa hết Cu mà không tạo khí => pư Cu với Fe³⁺ => tạo Fe²⁺, Cu²⁺

n_NO =0,28

Bảo toản e: 2n_Fe + 2n_Cu = 3n_NO

=> n_Fe = (3.0,28 – 2.0,13):2 = 0,29

=> m_Fe = 16,24 g

Đáp án :A