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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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Đọc tiếp

body 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, ::-webkit-file-upload-button, body .sweezy-custom-cursor-hover { cursor: url("data:image/png;base64,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") 6 3, pointer !important; } html, body, body select, body .sweezy-custom-cursor-default-hover { cursor: url("data:image/png;base64,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") 1 1, auto !important; }tính thể tích của 16g khí O2 ở đktc

0
23 tháng 4 2022

\(n_{Al}=\dfrac{4,5}{27}=\dfrac{1}{6}mol\)

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

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

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

=> Al dư

\(m_{Al\left(dư\right)}=\left(\dfrac{1}{6}-0,1\right).27=1,8g\)               

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

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

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

 

 

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)

21 tháng 11 2021

a) PTHH: Fe + H2SO===> FeSO+ H2

b) Ta có: nFe 1456=0,25(mol)

Theo PTHH, nH2SO4 = nFe = 0,25 (mol)

=> mH2SO4 = 0,25 x 98 = 24,5 (gam)

c) Theo PTHH, nH2 = nFe = 0,25 (mol)

=> VH2(đktc) = 0,25 x 22,4 = 5,6 (l)

d) Theo PTHH, nFeSO4 = nFe = 0,25 (mol)

=> mFeSO4(tạo thành) = 0,25 x 152 = 38 (gam)

9 tháng 5 2022

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

$b\big)$

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

Theo PT: $n_{H_2SO_4}=1,5n_{Al}=0,15(mol)$

$\to m_{dd\,H_2SO_4}=\dfrac{0,15.98}{30\%}=49(g)$

$c\big)$

Theo PT: $n_{H_2}=0,15(mol);n_{Al_2(SO_4)_3}=0,05(mol)$

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

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

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

\(\left\{{}\begin{matrix}n_{Al}=x\left(mol\right)\\n_{Ag}=y\left(mol\right)\end{matrix}\right.\Rightarrow27x+108y=4,2\left(1\right)\)

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

0,1                           0,05           0,15

\(\Rightarrow m_{Al}=0,1\cdot27=2,7g\)

\(\Rightarrow m_{Ag}=4,2-2,7=1,5g\)

a)\(\%m_{Al}=\dfrac{2,7}{4,2}\cdot100\%=64,28\%\)

\(\%m_{Ag}=100\%-64,28\%=35,72\%\)

b)\(m_{muối}=0,05\cdot342=17,1g\)