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

Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\) \(\Rightarrow n_{H_2SO_4}=n_{H_2}=0,3\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}m_{H_2SO_4}=0,3\cdot98=29,4\left(g\right)\\V_{H_2}=0,3\cdot22,4=6,72\left(l\right)\end{matrix}\right.\)

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

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

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

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

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

0,1         0,15                                0,15   ( mol )

\(V_{H_2}=0,15.22,4=3,36l\)

\(m_{H_2SO_4}=0,15.98=14,7g\)

nAl = 5.4/27 = 0.2 mol 

2Al + 3H2SO4 => Al2(SO4)3 + 3H2 

0.2______________0.1________0.3 

VH2 = 0.3*22.4 = 6.72 (l) 

mAl2(SO4)3 = 0.1*342 = 34.2 g 

Gọi x, y lần lượt là số mol Al, Fe

\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ n_{H_2}=\dfrac{0,56}{22,4}=0,025\left(mol\right)\\ Tacó:\left\{{}\begin{matrix}27x+56y=0,83\\1,5x+y=0,025\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=0,01\\y=0,01\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}m_{Al}=0,27\left(g\right)\\m_{Fe}=0,56\left(g\right)\end{matrix}\right.\)

\(\text{Em sửa 7,1 thành 8,1g nha}\\ a/ 2Al+3H_2SO_4 \rightarrow Al_2(SO_4)_3+3H_2\\ b/ \\ n_{Al}=0,3mol\\ n_{H_2SO_4}=\frac{3}{2}n_{Al}=\frac{3}{2}.0,3=0,45mol V_{H_2SO_4}=\frac{0,45}{0,05}=9M\\ c/ \\ V_{H_2}=0,45.22,4=10,08l\)

a)\(n_{Al}=\dfrac{7,1}{27}\approx0,26\left(mol\right)\)

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

          0,26----->0,39-------->0,13------->0,39 (mol)

b)\(V_{H_2SO_4}=0,39\cdot22,4=8,736\left(l\right)\)

c)\(V_{H_2}=0,39\cdot22,4=8,736\left(l\right)\)

\(a.2Al+3H2SO4\rightarrow Al2\left(SO4\right)3+3H2\)

b)\(n_{Al}=\dfrac{7,1}{27}=\dfrac{71}{270}\left(mol\right)\)

\(n_{H2SO4}=\dfrac{3}{2}n_{Al}=\dfrac{71}{180}\left(mol\right)\)

\(\Rightarrow V=\dfrac{71}{180}:0,05=\dfrac{71}{9}\left(l\right)\)

c)\(n_{H2}=\dfrac{3}{2}n_{Al}=\dfrac{71}{180}\left(mol\right)\)

\(V_{H2}=\dfrac{71}{180}.22,4=\dfrac{476}{225}\left(l\right)\)

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

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

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

\(m_{Al_2\left(SO_4\right)_3}=0,1.342=34,2g\)

\(V_{H_2}=0,3.22,4=6,72l\)

2Al + 3H₂SO₄ -> Al₂(SO₄)₃ + 3H₂

n_Al=\(\dfrac{10,8}{27}=0,4\left(mol\right)\)

Theo PTHH ta có:

\(\dfrac{3}{2}\)n_Al=n_H2SO4=n_H2=0,6(mol)

\(\dfrac{1}{2}\)n_Al=n_Al2(SO4)3=0,2(mol)

m_H2SO4=98.0,6=58,8(g)

V_H2=0,6.22,4=13,44(lít)

C1:m_Al2(SO4)3=0,2.342=68,4(g)

C2:

Áp dụng định luật BTKL ta có:

m_Al+m_H2SO4=m_Al2(SO4)3+m_H2

=>m_Al2(SO4)3=m_Al+m_H2SO4-m_H2

=10,8+58,8-0,6.2=68,4(g)

PTHH: \(2Al\left(OH\right)_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+6H_2O\)

Ta có: \(n_{Al\left(OH\right)_3}=\dfrac{7,8}{78}=0,1\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{H_2SO_4}=0,15mol\\n_{Al_2\left(SO_4\right)_3}=0,05mol\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{H_2SO_4}=0,15\cdot98=14,7\left(g\right)\\m_{Al_2\left(SO_4\right)_3}=0,05\cdot342=17,1\left(g\right)\end{matrix}\right.\)