\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\\ pthh:Zn+2HCl\rightarrow ZnCl_2+H_2\) 
          0,1           0,2            0,1      0,1 
\(m_{HCl}=0,2.36,5=7,3\left(g\right)\\ V_{H_2}=0,1.22,4=2,24l\\ m_{\text{dd}}=6,5+200-\left(0,1.2\right)=206,3g\)  
bài 2 :
\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ pthh:Mg+2HCl\rightarrow MgCl_2+H_2\) 
          0,2             0,4       0,2              0,2 
\(m_{HCl}=0,4.36,5=14,6g\\ V_{H_2}=0,2.22,4=4,48l\\ m\text{dd}=4,8+200-0,4=204,4g\\ C\%=\dfrac{0,2.136}{204,4}.100\%=13,3\%\)

\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)

PTHH :

\(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)

0,2     0,4            0,2        0,2 

\(a,V_{H_2}=0,2.22,4=4,48\left(l\right)\)

\(b,m_{HCl}=0,4.36,5=14,6\left(g\right)\)

\(m_{ddHCl}=\dfrac{14,6.100}{10}=146\left(g\right)\)

\(c,m_{MgCl_2}=0,2.95=19\left(g\right)\)

\(m_{ddMgCl_2}=4,8+146-\left(0,2.2\right)=150,4\left(g\right)\)

\(C\%_{MgCl_2}=\dfrac{19}{150,4}.100\%\approx12,63\%\)

2. 

\(n_K=\dfrac{7,8}{39}=0,2\left(mol\right)\)

\(2K+2H_2O\rightarrow2KOH+H_2\uparrow\)

0,2                     0,2         0,1 

\(m_{KOH}=0,2.56=11,2\left(g\right)\)

\(m_{ddKOH}=7,8+100-\left(0,1.2\right)=107,6\left(g\right)\)

\(C\%=\dfrac{11,2}{107,6}.100\%\approx10,4\%\)

\(a,n_{Fe}=\dfrac{5,6}{56}=0,1mol\\ Fe+2HCl\rightarrow FeCl_2+H_2\)

0,1      0,2             0,1          0,1

\(V_{H_2}=0,1.22,4=2,24l\\ b)C_{\%HCl}=\dfrac{0,2.36,5}{200}\cdot100=3,65g\)     

\(m_{NaOH}=\dfrac{200\cdot8}{100}=16\left(g\right)\Rightarrow n_{NaOH}=\dfrac{16}{40}=0,4mol\)

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

 0,4           0,4         0,4        0,4

a)\(m_{HCl}=0,4\cdot36,5=14,6\left(g\right)\) 

 \(\Rightarrow m_{ddHCl}=\dfrac{14,6}{7,3}\cdot100=200\left(g\right)\)

b)\(m_{NaCl}=0,4\cdot58,5=23,4\left(g\right)\)

   \(m_{H_2O}=0,4\cdot18=7,2\left(g\right)\)

   \(m_{ddsau}=200+200-7,2=392,8\left(g\right)\)

   \(\Rightarrow C\%=\dfrac{23,4}{392,8}\cdot100=5,96\%\)

c) \(n_{SO_2}=\dfrac{6,72}{22,4}=0,3mol\)

     \(2NaOH+SO_2\rightarrow Na_2SO_4+H_2O\)

         0,4          0,3         0,3              0,3

     \(m_{Na_2SO_4}=0,3\cdot142=42,6\left(g\right)\)

https://hoc24.vn/cau-hoi/cho-400-g-dung-dich-naoh-30-tac-dung-vua-het-voi-100-g-dung-dich-hcl-tinh-a-nong-do-muoi-thu-duoc-sau-phan-ungb-tinh-nong-do-axit-hcl-bie.7974814552205

Tui trả lời rùi nghen

Bài 3 : 

\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)

Pt : \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2|\)

        2            3                   1               3

       0,1       0,15                 0,05       0,15

a) \(n_{H2}=\dfrac{0,1.3}{2}=0,15\left(mol\right)\)

\(m_{H2}=0,15.2=0,3\left(g\right)\)

\(V_{H2\left(dktc\right)}=0,15.22,4=3,36\left(l\right)\)

b) \(n_{H2SO4}=\dfrac{0,1.3}{2}=0,15\left(mol\right)\)

⇒ \(m=0,15.98=14,7\left(g\right)\)

\(C_{ddH2SO4}=\dfrac{14,7.100}{200}=7,35\)⁰/₀

c) \(n_{Al2\left(SO4\right)3}=\dfrac{0,15.1}{3}=0,05\left(mol\right)\)

⇒ \(m_{Al2\left(SO4\right)3}=0,05.342=17,1\left(g\right)\)

\(m_{ddspu}=2,7+200-0,3=302,4\left(g\right)\)

\(C_{Al2\left(SO4\right)3}=\dfrac{17,1.100}{302,4}=5,65\)⁰/₀

 Chúc bạn học tốt

a, PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)

\(MgCl_2+2NaOH\rightarrow2NaCl+Mg\left(OH\right)_{2\downarrow}\)

\(Mg\left(OH\right)_2\underrightarrow{t^o}MgO+H_2O\)

b, Ta có: \(n_{Mg}=\dfrac{9,6}{24}=0,4\left(mol\right)\)

Theo PT: \(n_{HCl}=2n_{Mg}=0,8\left(mol\right)\)

\(\Rightarrow C_{M_{HCl}}=\dfrac{0,8}{0,2}=4\left(M\right)\)

c, Theo PT: \(n_{MgO}=n_{Mg}=0,4\left(mol\right)\)

\(\Rightarrow m_{MgO}=0,4.40=16\left(g\right)\)

ZN + 2HCl -> ZnCl2 + H2

a) nZn = 0.3 mol

nH2 = nZn = 0.3 mol

VH2 = 0.3. 22.4 = 6.72 lít

b) nH2 = 0.3 mol

n Fe2O3 = 0.12 mol

tỉ lệ  

nH2/3 < nFe2O3/ 1

=> Fe2O3 dư

nFe = 2/3 nH2 =0.1 mol

=> mFe = 0.1. 56 = 5.6 gam

bài 2 và 3 dễ rồi chắc bạn vẫn có thể làm được

Bài 1:hòa tan 19.5g kẽm bằng đ axit clohiddric

a) thể tích H2 sinh ra (dktc)

b) nếu dùng VH2 trên để khử 19,2g sắt III oxit thì thu được bao nhiêu g sắt?

Bài 2: cho 2,4g Mg tác dụng vừa đủ với m gam dd HCl 20%. Biết D=1,1g/ml