a)\(n_{H_2}=\dfrac{19,832}{24,79}=0,8\left(mol\right)\)

\(PTHH:Zn+H_2SO_4\xrightarrow[]{}ZnSO_4+H_2\)

tỉ lệ       :1        1               1              1

số mol  :0,8      0,8            0,8          0,8

\(m_{Zn}=0,8.65=52\left(g\right)\)

b)\(m_{H_2SO_4}=0.8.98=78,4\left(g\right)\)

c)\(m_{ZnSO_4}=0,8.161=128\left(g\right)\)

a)Số mol của Al là:

\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)

\(PTHH:2Al+6HCl\xrightarrow[]{}2AlCl_3+3H_2\)

tỉ lệ       : 2         6            2               3    (mol)

số mol  : 0,2      0,6         0,2            0,3 (mol)

Thể tích khí H₂ thu được là:

\(V_{H_2}=0,3.24,79=7,437\left(l\right)\)

b)Khối lượng muối thu được là:

\(m_{AlCl_3}=0,2.133,5=26,7\left(g\right)\)

c)\(PTHH:Fe_2O_3+3H_2\xrightarrow[]{t^0}2Fe+3H_2O\)

tỉ lệ           :1              3           2        3     (mol)

số mol      :0,1           0,3        0,2     0,3  (mol)

Khối lượng kim loại tạo thành là:

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

\(4.a/n_{Al}=\dfrac{5,4}{27}=0,2mol\\ 2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

0,2       0,3                  0,1                  0,3

\(V_{H_2}=0,3.24,79=7,437l\\ b/C_{\%H_2SO_4}=\dfrac{0,3.98}{150}\cdot100=19,6\%\\ c/m_{Al_2\left(SO_4\right)_3}=0,1.342=34,2g\)

\(5.a/n_{MgO}=\dfrac{4}{40}=0,1mol\\ MgO+2HCl\rightarrow MgCl_2+H_2O\)

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

\(C_{\%HCl}=\dfrac{0,2.36,5}{200}\cdot100=3,65\%\\ b/C_{\%MgCl_2}=\dfrac{0,1.95}{200+4}\cdot100=4,66\%\\ c/NaOH+HCl\rightarrow NaCl+H_2O\\ n_{NaOH}=n_{HCl}=0,2mol\\ V_{NaOH}=\dfrac{0,2}{1}=0,2l=200ml\)

\(a)n_{H_2}=\dfrac{14,874}{24,79}=0,6mol\\2 Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

0,4mol     0,6mol          0,2mol           0,6mol

\(m_{Al}=0,4.27=10,8g\\ b)m_{Al_2\left(SO_4\right)_3}=0,2.342=68,4g\\ c)m_{H_2SO_4}=0,6.98=58,8g\)

\(m_{H_2}=\dfrac{7,437}{24,79}=0,3mol\\ 2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

0,2        0,3               0,1                0,3

\(m_{Al}=0,2.27=5,4g\\ V_{ddH_2SO_4}=\dfrac{0,3}{0,5}=0,6l\\ C_{M_{Al_2\left(SO_4\right)_3}}=\dfrac{0,1}{0,6}=\dfrac{1}{6}\approx0,167M\)

a, \(n_{H_2}=\dfrac{3,7185}{24,79}=0,15\left(mol\right)\)

PT: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

Theo PT: \(n_{Fe}=n_{H_2}=0,15\left(mol\right)\Rightarrow m_{Fe}=0,15.56=8,4\left(g\right)\)

b, Theo PT: \(n_{FeSO_4}=n_{H_2}=0,15\left(mol\right)\Rightarrow m_{FeSO_4}=0,15.152=22,8\left(g\right)\)

c, \(n_{CuO}=\dfrac{24}{80}=0,3\left(mol\right)\)

PT: \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)

Xét tỉ lệ: \(\dfrac{0,3}{1}>\dfrac{0,15}{1}\), ta được CuO dư.

Theo PT: \(n_{Cu}=n_{CuO\left(pư\right)}=n_{H_2}=0,15\left(mol\right)\)

\(\Rightarrow n_{CuO\left(dư\right)}=0,3-0,15=0,15\left(mol\right)\)

Chất rắn thu được sau pư gồm Cu và CuO dư.

⇒ m _{chất rắn} = m_Cu + m_{CuO (dư)} = 0,15.64 + 0,15.80 = 21,6 (g)

Bài 8:

\(n_{H_2}=\dfrac{7,437}{24,79}=0,3\left(mol\right)\)

PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

_____0,2______0,6_____0,2____0,3 (mol)

a, \(m_{Al}=0,2.27=5,4\left(g\right)\)

b, \(C_{M_{HCl}}=\dfrac{0,6}{0,3}=2\left(M\right)\)

c, \(C_{M_{AlCl_3}}=\dfrac{0,2}{0,2}=1\left(M\right)\)

Bài 9:

Ta có: \(n_{H_2}=\dfrac{2,479}{24,79}=0,1\left(mol\right)\)

PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)

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

a, \(n_{Mg}=n_{H_2}=0,1\left(mol\right)\Rightarrow m_{Mg}=0,1.24=2,4\left(g\right)\)

\(\Rightarrow m_{MgO}=8,4-2,4=6\left(g\right)\)

b, \(n_{MgO}=\dfrac{6}{40}=0,15\left(mol\right)\)

Theo PT: \(n_{HCl}=2n_{Mg}+2n_{MgO}=0,5\left(mol\right)\)

\(\Rightarrow m_{ddHCl}=\dfrac{0,5.36,5}{3,65\%}==500\left(g\right)\)

\(a.2Al+3H_2SO_4->Al_2\left(SO_4\right)_3+3H_2\\ b.n_{Al}=1,5.0,5.0,03=0,0375mol\\ m_{Al}=0,0375.27=1,0125g\\ m_{Al_2\left(SO_4\right)_3}=342\cdot\dfrac{1}{3}\cdot0,03\cdot0,5=1,71g\\V_{H_2}=24,79.0,5.0,03=0,37185L\)

\(a/2Al+3H_2SO_4\xrightarrow[]{}Al_2\left(SO_4\right)_3+3H_2\\ b/n_{H_2SO_4}=0,5.0,03=0,015\left(mol\right)\\ n_{Al_2\left(SO_4\right)_3}=\dfrac{0,015}{3}=0,005\left(mol\right)\\ m_{Al_2\left(SO_4\right)_3}=0,005.342=1,71\left(g\right)\\ n_{Al}=0,005.2=0,01\left(mol\right)\\ m_{Al}=0,01.27=0,27\left(g\right)\\ c/n_{H_2}=0,005.3=0,015\left(mol\right)\\ V_{H_2}=0,015.24,79=0,37185\left(l\right)\)

\(a/2Al+3H_2SO_4\xrightarrow[]{}Al_2\left(SO_4\right)_3+3H_2\)

\(b/30ml=0,03l\\ n_{H_2SO_4}=0,5.0,03=0,0015\left(mol\right)\\ n_{Al}=\dfrac{0,0015.2}{3}=0,001\left(mol\right)\\ m_{Al}=0,001.27=0,027\left(g\right)\\ n_{Al_2\left(SO_4\right)_3}=\dfrac{0,0015}{2}=0,00075\left(mol\right)\\ m_{Al_2\left(SO_4\right)_3}=0,00075.342=0,2565\left(g\right)\)

\(c/n_{H_2}=\dfrac{0,0015.3}{3}=0,0015\left(mol\right)\\ V_{H_2}=0,0015.24,79=0,037185\left(l\right)\)