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1 tháng 11 2021

a. PTHH:

Al+H2SO4-->AlSO4+H2

b.Theo ĐLBTKL, ta có:

mAl+mH2SO4=mAl2SO4+mH2

=>mH2SO4=mAl2SO4+mH2-mAl

=171+3-27=147 (g)

19 tháng 4 2022

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

19 tháng 4 2022

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

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

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

b: \(n_{H2}=\dfrac{3.36}{22.4}=0.15\left(mol\right)\)

\(\Leftrightarrow n_{Al}=0.1\left(mol\right)\)

\(m_{Al}=n_{Al}\cdot M_{Al}=0.1\cdot27=2.7\left(g\right)\)

4 tháng 5 2022

Sai

 

 

24 tháng 1 2022

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

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

H2=\(\dfrac{3,36}{22,4}\)=0,15 mol

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

=>m H2SO4=0,15.98=14,7g

 

24 tháng 1 2022

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6 tháng 2 2022

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!

6 tháng 2 2022

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

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

 nH2=\(\dfrac{3,36}{22,4}\)=0,15 mol

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

=>m H2SO4=0,15.98=14,7g

Cho kim loại Al vào đ H2SO4 sau phản ứng thu được 3,36 lít khí đktc và muối nhôm sunfat và khí hidroa Viết PTHH xảy ra?b Tính khối lượng Al sau phản ứngc Tính khối lượng muối thu được và khối lượng axit đã phản ứngbody a, body button, body [type='button'], body input[type='reset'], body input[type='submit'], body [role="button"], ::-webkit-search-cancel-button, ::-webkit-search-decoration, ::-webkit-scrollbar-button, ...
Đọc tiếp

Cho kim loại Al vào đ H2SO4 sau phản ứng thu được 3,36 lít khí đktc và muối nhôm sunfat và khí hidro

a Viết PTHH xảy ra?

b Tính khối lượng Al sau phản ứng

c Tính khối lượng muối thu được và khối lượng axit đã phản ứng

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0
21 tháng 11 2021

`a)`

PTHH : `2Al + 3H_2SO_4 -> Al_2(SO_4)_3 + 3H_2`

`b)`

`n_{H_2} = (3,36)/(22,4) = 0,15` `mol`

`n_{H_2SO_4} = n_{H_2} = 0,15` `mol`

`m_{H_2SO_4} = 0,15 . 98 = 14,7` `gam`

`c)`

`n_{Al_2(SO_4)_3} = 1/3 . n_{H_2} = 0,05` `mol`

`m_{Al_2(SO_4)_3} = 0,05 . 342 = 17,1` `gam`

27 tháng 12 2020

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

Làm gộp các phần còn lại

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

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

 

 

25 tháng 11 2021

a. \(PTHH:2Al+3H_2SO_4--->Al_2\left(SO_4\right)_3+3H_2\)

b. Áp dụng định luật bảo toàn khối lượng, ta có:

\(m_{Al}+m_{H_2SO_4}=m_{Al_2\left(SO_4\right)_3}+m_{H_2}\)

\(\Leftrightarrow m_{Al_2\left(SO_4\right)_3}=m_{Al}+m_{H_2SO_4}-m_{H_2}=5,4+29,4-0,6=34,2\left(g\right)\)

25 tháng 11 2021

giúp mình câu C

25 tháng 11 2021

i