nK2SO3=15.8\158=0.1(mol)

K2SO3+H2SO4→K2SO4+SO2+H2O

0.1...........................0.1..........0.1

VSO2=0.1⋅22.4=2.24(l)

CMK2SO4=0.1\0.2=0.5(M)

câu A à bạn

\(n_{HCl}=0,3.2=0,6\left(mol\right)\\a, 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ b,n_{H_2}=\dfrac{3}{6}.0,6=0,3\left(mol\right)\\ V_{H_2\left(đktc\right)}=0,3.22,4=6,72\left(l\right)\\ c,n_{Al}=n_{AlCl_3}=\dfrac{2}{6}.0,6=0,2\left(mol\right)\\ m_{Al}=0,2.27=5,4\left(g\right)\\ d,V_{ddAlCl_3}=V_{ddHCl}=0,3\left(l\right)\\ C_{MddHCl}=\dfrac{0,2}{0,3}=\dfrac{2}{3}\left(M\right)\)

đổi 200 ml  = 0,02 l

a) PTHH : Fe + HCl -> FeCl₂ + H₂

b) \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

\(=>V_{H_2}=0,1.22,4=2,24\left(l\right)\)

\(C_{HCl}=\dfrac{n}{V}=\dfrac{0,1}{0,02}=5\left(M\right)\)

Đông Hải làm câu 1 rồi thì tui làm phần còn lại

Câu 2:

\(a,PTHH:CuO+H_2\underrightarrow{t^o}Cu+H_2O\\ b,n_{CuO}=\dfrac{m}{M}=\dfrac{8}{80}=0,1\left(mol\right)\\ Theo.PTHH:n_{Cu}=n_{H_2}=n_{CuO}=0,1\left(mol\right)\\ V_{H_2\left(đktc\right)}=n.22,4=0,1.22,4=2,24\left(l\right)\\ c,m_{Cu}=n.M=0,1.64=6,4\left(g\right)\\ d,PTHH:2H_2+O_2\underrightarrow{t^o}2H_2O\left(2\right)\\ Theo.PTHH\left(2\right):n_{H_2O}=n_{H_2}=0,1\left(mol\right)\)

\(m_{H_2O}=n.M=0,1.18=1,8\left(g\right)\)

\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\\ pthh:Zn+2HCl\rightarrow ZnCl_2+H_2\) 
          0,2        0,4                      0,2 
\(V_{H_2}=0,2.22,4=4,48\left(l\right)\\ C\%_{HCl}=\dfrac{0,4.36,5}{200}.100\%=7,3\%\)

nZn = 19,5 / 65 = 0,3 (mol)

Zn + 2HCl -->ZnCl2 + H2

0,3    0,6          0,3       0,3   (mol)

VH2 =0,3.22,4 = 6,72 (l)

V = 200ml = 0,2 (l)

CMHCl = n/V = 0,6 / 0,2 = 3M

Ta có: \(n_{CaSO_3}=\dfrac{12,1}{120}=\dfrac{121}{1200}\left(mol\right)\)

a. \(PTHH:CaSO_3+H_2SO_4--->CaSO_4+SO_2+H_2O\)

b. Theo PT: \(n_{SO_2}=n_{CaSO_3}=\dfrac{121}{1200}\left(mol\right)\)

\(\Rightarrow V_{SO_2}=\dfrac{121}{1200}.22,4=\dfrac{847}{375}\left(lít\right)\)

c. Theo PT: \(n_{H_2SO_4}=n_{SO_2}=\dfrac{121}{1200}\left(mol\right)\)

\(\Rightarrow C_{M_{H_2SO_4}}=\dfrac{\dfrac{121}{1200}}{\dfrac{300}{1000}}\approx0,336M\)

(Mik nghĩ là đề sai nhé.)

\(a.MgO+2HCl\rightarrow MgCl_2+H_2O\)

\(b.n_{MgO}=\dfrac{16}{40}=0,04mol\)

\(\rightarrow n_{HCl}=0,04.2=0,08mol\)

\(C_{M_{HCl}}=\dfrac{0,08}{0,15}=0,53M\)

\(c.m_{MgCl_2}=0,04.95=3,8g\)

Ta có: \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

a. PTHH: Fe + H₂SO₄ ---> FeSO₄ + H₂

b. Theo PT: \(n_{H_2}=n_{Fe}=0,1\left(mol\right)\)

=> \(V_{H_2}=0,1.22,4=2,24\left(lít\right)\)

c. Theo PT: \(n_{H_2SO_4}=n_{Fe}=0,1\left(mol\right)\)

=> \(m_{H_2SO_4}=0,1.98=9,8\left(g\right)\)

=> \(C_{\%_{H_2SO_4}}=\dfrac{9,8}{100}.100\%=9,8\%\)