Ta có : \(m_{H_2SO_4}=14\%.200=28\left(g\right)\)

\(\Rightarrow n_{H_2SO_4}=\frac{28}{98}=\frac{2}{7}\left(mol\right)\)

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

(mol)                2/7             2/21

\(\Rightarrow m_{Al_2\left(SO_4\right)_3}=\frac{2}{21}.342=\frac{228}{7}\left(g\right)\)

Sửa: \(14,7\%\)

\(n_{H_2SO_4}=\dfrac{200.14,7\%}{100\%.98}=0,3(mol)\\ a,PTHH:2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2\\ \Rightarrow n_{H_2}=0,3(mol)\\ \Rightarrow V_{H_2}=0,3.22,4=6,72(l)\\ b,n_{Al}=\dfrac{2}{3}n_{H_2SO_4}=0,2(mol)\\ \Rightarrow m_{Al}=0,2.27=5,4(g)\\ c,n_{Al_2(SO_4)_3}=\dfrac{1}{2}n_{Al}=0,1(mol)\\ \Rightarrow C\%_{Al_2(SO_4)_3}=\dfrac{0,1.342}{200+5,4-0,3.2}.100\%=16,7\%\\ c,m_{Al_2(SO_4)_3}=0,1.342=34,2(g)\)

\(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right);n_{H_2SO_4}=\dfrac{98}{98}=1\left(mol\right)\)

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

Xét tỉ lệ \(\dfrac{0,4}{2}< \dfrac{1}{3}\) => Al hết, H₂SO₄ dư

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

_____0,4--->0,6------------------------->0,6

=> n_{H2SO4 dư} = 1-0,6=0,4(mol)

=> V_H2 = 0,6.22,4 = 13,44(l)

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

Ta có: \(C_{\%_{H_2SO_4}}=\dfrac{m_{H_2SO_4}}{200}.100\%=14,7\%\)

=> \(m_{H_2SO_4}=29,4\left(g\right)\)

=> \(n_{H_2SO_4}=\dfrac{29,4}{98}=0,3\left(mol\right)\)

Theo PT: \(n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{3}.n_{H_2SO_4}=\dfrac{1}{3}.0,3=0,1\left(mol\right)\)

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

Ta có: \(n_{H_2}=\dfrac{74,37}{24,79}=3\left(mol\right)\)

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

a, \(n_{Al}=\dfrac{2}{3}n_{H_2}=2\left(mol\right)\)

\(\Rightarrow m_{Al}=2.27=54\left(g\right)\)

b, \(n_{H_2SO_4}=n_{H_2}=3\left(mol\right)\)

\(\Rightarrow m_{H_2SO_4}=3.98=294\left(g\right)\)

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

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

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

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

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

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

\(Đặt:n_{Al}=a\left(mol\right);n_{Mg}=b\left(mol\right)\left(a,b>0\right)\\ PTHH:2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ Mg+H_2SO_4\rightarrow MgSO_4+H_2\\ \Rightarrow\left\{{}\begin{matrix}27a+24b=4,5\\1,5.22,4a+22,4b=5,04\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,075\end{matrix}\right.\\ \Rightarrow\%m_{Al}=\dfrac{0,1.27}{4,5}.100=60\%\\ \Rightarrow\%m_{Mg}=100\%-60\%=40\%\)