a) 2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂

b) n_Al = \(\frac{40,5}{27}=1,5\left(mol\right)\)

Từ PT \(\Rightarrow n_{H_2SO_4}=2,25\left(mol\right);n_{Al_2\left(SO_4\right)_3}=0,75\left(mol\right);n_{H_2}=2,25\left(mol\right)\)

\(\Rightarrow m_{H_2SO_4}=2,25.98=220,5\left(g\right)\)

c) \(m_{Al_2\left(SO_4\right)_3}=0,75.342=256,5\left(g\right)\)

d) đktc : \(V_{H_2}=22,4.2,25=50,4\left(l\right)\)

a)     2Al + 3H2SO4 → Al2(SO4)3 + 3H2  (1)

b) nAl = 40,5 : 27 = 1,5 mol

Từ pt(1) suy ra : nH2SO4 = \(\frac{3}{2}nAl\) = \(\frac{3}{2}.1,5=2,25mol\)

Khối lượng H2SO4 là : mH2SO4 = 2,25 . 98 = 220,5 g

c) Từ pt(1) => nAl2(SO4)3 = \(\frac{1}{2}nAl=\frac{1}{2}.1,5=0,75mol\)

=> mAl2(SO4)3 = 0,75 . 342 = 256,5 g

d) Từ pt(1) => nH2 = nH2SO4 = 2,25 mol

Thể tích khí H2 là : V_H2=2,25 . 22,4 = 50,4 lit

Câu 7 : 

\(n_{H2}=\dfrac{17,92}{22,4}=0,8\left(mol\right)\)

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

         2            3                 1              3

       \(\dfrac{8}{15}\)         0,8                \(\dfrac{4}{15}\)         0,8

\(n_{H2SO4}=\dfrac{0,8.3}{3}=0,8\left(mol\right)\)

⇒ \(m_{H2SO4}=0,8.98=78,4\left(g\right)\)

\(n_{Al2\left(SO4\right)3}=\dfrac{0,8.1}{3}=\dfrac{4}{15}\left(mol\right)\)

⇒ \(m_{Al2\left(SO4\right)3}=\dfrac{4}{15}.342=91,2\left(g\right)\)

\(n_{Al}=\dfrac{0,8.2}{3}=\dfrac{8}{15}\left(mol\right)\)

⇒ \(m_{Al}=\dfrac{8}{15}.27=14,4\left(g\right)\)

 Chúc bạn học tốt

bài 8

Al₂O₃+6HNO₃->2Al(NO₃)₃+3H₂

\(\dfrac{2}{15}\)------------0,8-------\(\dfrac{4}{15}\)----------0,4 mol

n _H2O=\(\dfrac{7,2}{18}\)=0,4 mol

=>m _Al2O3=\(\dfrac{2}{15}\).102=13,6g

=>m _HNO3=0,8.63=50,4g

=>m _Al(NO3)3=\(\dfrac{4}{15}\).213=56,8g

Bài 7

2Al+3H₂SO₄->Al₂(SO₄)₃+3H₂

\(\dfrac{8}{15}\)------0,5-----------\(\dfrac{4}{15}\)--------0,8

n H2=\(\dfrac{17,92}{22,4}\)=0,8 mol

=>m Al= \(\dfrac{8}{15}\).27=14,4g

=>m HCl=0,5.36,5=18,25g

=>m Al2(SO4)3=\(\dfrac{4}{15}\).342=91,2g

Phương trình hóa học của phản ứng:

So sánh tỉ lệ  → Vậy Al2O3 dư

mAl2O3 (dư) = 60 - 17 = 43(g)

a.b.\(n_{Al}=\dfrac{m_{Al}}{M_{Al}}=\dfrac{5,4}{27}=0,2mol\)

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

0,2                                              0,3    ( mol )

\(V_{H_2}=n_{H_2}.22,4=0,3.22,4=6,72l\)

c.\(n_{CuO}=\dfrac{m_{CuO}}{M_{CuO}}=\dfrac{32}{80}=0,4mol\)

\(CuO+H_2\rightarrow Cu+H_2O\)

 0,4  <  0,3                          ( mol )

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

\(m_A=m_{CuO\left(du\right)}+m_{Cu}=\left[\left(0,4-0,3\right).80\right]+\left(0,3.64\right)=8+19,2=27,2g\)

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

0,1----------------------0,075----0,15

n H2=0,15 mol

=>mAl=0,1.27=2,7g

=>m Al2(SO4)3=0,075.342=25,65g

a) PTHH: \(2Al+3H_2SO_2\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

b) \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

\(n_{Al}=\dfrac{2}{3}.0,15=0,1\left(mol\right)\)

\(m_{Al}=0,1.27=2,7\left(g\right)\)

c) \(n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{2}.0,1=0,05\left(mol\right)\)

\(m_{Al_2\left(SO_4\right)_3}=0,05.342=17,1\left(g\right)\)

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

0,1----------------------0,075----0,15

n H2=0,15 mol

=>mAl=0,1.27=2,7g

=>m Al2(SO4)3=0,075.342=25,65g

a) $2Al + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2$
b)

n Al = 2,7/27 = 0,1(mol)

Theo PTHH :

n H2SO4 = 3/2 n Al = 0,15(mol)

=> C% H2SO4 = 0,15.98/441  .100% = 3,33%

c)

Theo PTHH :

n H2 = 3/2 n Al = 0,15(mol)

=> V H2 = 0,15.22,4 = 3,36(lít)