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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 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, ...
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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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0
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)

12 tháng 9 2023

\(\left(a\right)2Al+3H_2O\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ \left(b\right)n_{H_2}=\dfrac{13,44}{22,4}=0,6mol\\ n_{Al}=\dfrac{0,6.2}{3}=0,4mol\\ m_{Al}=0,4.27=10,8g\\ \left(c\right)n_{O_2}=\dfrac{4,8}{32}=0,15mol\\ 4Al+3O_2\underrightarrow{t^0}2Al_2O_3\\ \Rightarrow\dfrac{0,4}{4}>\dfrac{0,15}{3}\Rightarrow Al.dư\\ n_{Al_2O_3}=\dfrac{0,15.2}{3}=0,1mol\\ m_{oxit}=m_{Al_2O_3}=0,1.102=10,2g\)

a: \(2Al+3H_2SO_4\rightarrow1Al_2\left(SO_4\right)_3+3H_2\uparrow\)

    0,4           0,6                0,2             0,6

b: \(n_{H_2}=\dfrac{13.44}{22.4}=0.6\left(mol\right)\)

=>\(n_{Al}=0.4\left(mol\right)\)

\(m_{Al}=0.4\cdot27=10.8\left(g\right)\)

c: \(4Al+3O_2\rightarrow2Al_2O_3\)

0,4                      0,2

\(m_{Al_2O_3}=0.2\left(27\cdot2+16\cdot3\right)=0.2\cdot102=20.4\left(g\right)\)

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

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

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

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

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

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

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

25 tháng 11 2023

Ta có: \(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

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

a, \(n_{Al}=\dfrac{2}{3}n_{H_2}=\dfrac{1}{6}\left(mol\right)\Rightarrow m_{Al}=\dfrac{1}{6}.27=4,5\left(g\right)\)

b, \(n_{H_2SO_4}=n_{H_2}=0,25\left(mol\right)\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,25}{0,2}=1,25\left(M\right)\)

c, \(n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{3}n_{H_2}=\dfrac{1}{12}\left(mol\right)\)

\(\Rightarrow m_{Al_2\left(SO_4\right)_3}=\dfrac{1}{12}.342=28,5\left(g\right)\)

25 tháng 12 2022

a)

$4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3$
b)

$n_{Al} = \dfrac{8,1}{27} = 0,3(mol) ; n_{O_2} = \dfrac{6,72}{22,4} = 0,3(mol)$

Ta thấy : 

$n_{Al} : 4 < n_{O_2} : 3$ nên $O_2$ dư

$n_{Al_2O_3} = \dfrac{1}{2}n_{Al} = 0,15(mol)$
$m_{Al_2O_3} = 0,15.102 = 15,3(gam)$

c) $n_{O_2\ pư} = \dfrac{3}{4}n_{Al} = 0,225(mol)$
$\Rightarrow m_{O_2\ dư} = (0,3 - 0,225).32 = 2,4(gam)$