Sửa đề thành 0,54 gam Al cho số mol đẹp bạn nhé!

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

Ta có: \(n_{Al}=\dfrac{0,54}{27}=0,02\left(mol\right)\)

Theo PT: \(\left\{{}\begin{matrix}n_{H_2SO_4}=n_{H_2}=\dfrac{3}{2}n_{Al}=0,03\left(mol\right)\\n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{2}n_{Al}=0,01\left(mol\right)\end{matrix}\right.\)

a, Ta có: \(n_{H_2}=0,03.22,4=0,672\left(l\right)\)

b, \(m_{H_2SO_4}=0,03.98=2,94\left(g\right)\)

c, \(m_{Al_2\left(SO_4\right)_3}=0,01.342=3,42\left(g\right)\)

Bạn tham khảo nhé!

\(n_{H_2}=\dfrac{6}{2}=3\left(mol\right)\)

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

\(2..............3.....................1.........3\)

\(m_{Al}=2\cdot27=54\left(g\right)\)

\(m_{H_2SO_4}=3\cdot98=294\left(g\right)\)

\(m_{Al_2\left(SO_4\right)_3}=1\cdot342=342\left(g\right)\)

a) $n_{H_2} = \dfrac{6}{2} = 3(mol)$

2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂

2............3.................1...............3..............(mol)

$m_{Al} = 2.27 = 54(gam)$

$m_{H_2SO_4} = 3.98 = 294(gam)$

b)

$m_{Al_2(SO_4)_3} = 1.342 = 342(gam)$

\(n_{Al}=\dfrac{1,35}{27}=0,05\left(mol\right)\\ 2Al+3H_2SO_4\rightarrow2Al_2\left(SO_4\right)_3+3H_2\\ n_{H_2SO_4}=n_{H_2}=\dfrac{3}{2}.0,05=0,075\left(mol\right)\\ n_{Al_2\left(SO_4\right)_3}=\dfrac{0,05}{2}=0,025\left(mol\right)\\ a,m_{Al_2\left(SO_4\right)_3}=342.0,025=8,55\left(g\right)\\ b,V_{H_2\left(đktc\right)}=0,075.22,4=1,68\left(l\right)\\ c,m_{H_2SO_4}=0,075.98=7,35\left(g\right)\)

\(n_{Al}=\dfrac{1,35}{27}=0,05mol\)

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

0,05      0,075        0,025           0,075

\(m_{Al_2\left(SO_4\right)_3}=0,025\cdot342=8,55g\)

\(V_{H_2}=0,075\cdot22,4=1,68l\)

\(m_{H_2SO_4}=0,075\cdot98=7,35g\)

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

Làm gộp các phần còn lại

Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}n_{Al_2\left(SO_4\right)_3}=0,1mol\\n_{H_2SO_4}=n_{H_2}=0,3mol\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}V_{H_2}=0,3\cdot22,4=6,72\left(l\right)\\m_{Al_2\left(SO_4\right)_3}=0,1\cdot342=34,2\left(g\right)\\m_{H_2SO_4}=0,3\cdot98=29,4\left(g\right)\end{matrix}\right.\) 

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

b) nH₂ =\(\dfrac{26,88}{22,4}\)=1,2 mol 

Theo tỉ lệ phản ứng => nAl phản ứng = \(\dfrac{nH_2.2}{3}\)= 0,8 mol

=> mAl phản ứng = 0,8.27= 21,6 gam

c) nAl₂(SO₄)₃ = 1/2 nAl = 0,4 mol

=> m Al₂(SO₄)₃ = 0,4. 342 = 136,8 gam

a) 

b) mAl=21,6gmAl=21,6g

c) mAl2(SO4)3=136,8gmAl2(SO4)3=136,8g

Giải thích các bước giải:

a) Phương trình hoá học:

b) Số mol  sinh ra sau phản ứng:

nAl=23nH2=23⋅1,2=0,8molnAl=23nH2=23⋅1,2=0,8mol

Khối lượng  tham gia phản ứng:

mAl=nAl.MAl=0,8.27=21,6gmAl=nAl.MAl=0,8.27=21,6g

c) Dựa vào phương trình hóa học ta được:

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

a

PTHH:

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

0,2--->0,6----->0,2------->0,3

b

\(C\%_{dd.HCl.đã.dùng}=\dfrac{0,6.36,5.100\%}{200}=10,95\%\)

c

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

2Al+3H₂SO₄->Al₂(SO4)₃+3H₂

0,1-------0,15---------------------0,15 mol

n _H2=\(\dfrac{3,36}{22,4}\)=0,15 mol

=>m _Al=0,1.27=2,7g

=>m _H2SO4=0,15.98=14,7g

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

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

              0,1        0,15             0,05         0,15 

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

c . \(m_{H_2SO_4}=0,15.98=14,7\left(g\right)\)

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

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

Theo PT: \(n_{Al}=\dfrac{2}{3}n_{H_2}=0,1\left(mol\right)\)

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

c, Theo PT: \(n_{H_2SO_4}=n_{H_2}=0,15\left(mol\right)\)

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

Bạn tham khảo nhé!

2Al+3H₂SO₄->Al₂(SO₄)₃+3H₂

0,1----0,15----------------------0,15 mol

 n_H2=\(\dfrac{3,36}{22,4}\)=0,15 mol

=>m _Al=0,1.27=2,7g

=>m _H2SO4=0,15.98=14,7g

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

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

             0,1         0,15            0,05          0,15           ( mol )

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

c. \(m_{H_2SO_4}=0,15.98=14,7\left(g\right)\)