\(n_{Al}=\dfrac{8.1}{27}=0.3\left(mol\right)\)

\(n_{H_2SO_4}=\dfrac{29.4}{98}=0.3\left(mol\right)\)

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

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

\(0.3..........0.3\)

\(LTL:\dfrac{0.3}{2}>\dfrac{0.3}{3}\Rightarrow Aldư\)

\(V_{H_2}=0.3\cdot22.4=6.72\left(l\right)\)

\(m_{Al\left(dư\right)}=\left(0.3-0.2\right)\cdot27=2.7\left(g\right)\)

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

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

\(n_{H_2SO_4}=\dfrac{30}{98}=\dfrac{15}{49}\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,2}{2}< \dfrac{\dfrac{15}{49}}{3}\) , ta được H₂SO₄ dư.

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

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

c, Sau phản ứng, trong cốc có H₂SO₄ dư và Al₂(SO₄)₃.

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

\(\Rightarrow n_{H_2SO_4\left(dư\right)}=\dfrac{15}{49}-0,3\approx0,006\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}m_{H_2SO_4\left(dư\right)}=0,006.98=0,588\left(g\right)\\m_{Al_2\left(SO_4\right)_3}=0,1.342=34,2\left(g\right)\end{matrix}\right.\)

Bạn tham khảo nhé!

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nAl=0.2(mol)

nH2SO4=0.4(mol)

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

Theo pthh,nH2SO4=3/2 nAl

Theo bài ra,nH2SO4=2nAl

-> H2SO4 dư tính theo  Al

nH2SO4 phản ứng:0.2*3/2=0.3(mol)

nH2SO4 dư=0.40.3=0.1(mol)

mH2SO4 dư:0.1*98=9.8(g)

Theo pthh,nH2=3/2 nAl->nH2=0.3(mol)

VH2=0.3*22.4=6.72(l)

nAl=0.2(mol)

nH2SO4=0.4(mol)

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

Theo pthh,nH2SO4=3/2 nAl

Theo bài ra,nH2SO4=2nAl

-> H2SO4 dư tính theo  Al

nH2SO4 phản ứng:0.2*3/2=0.3(mol)

nH2SO4 dư=0.40.3=0.1(mol)

mH2SO4 dư:0.1*98=9.8(g)

Theo pthh,nH2=3/2 nAl->nH2=0.3(mol)

VH2=0.3*22.4=6.72(l)

a) PTHH: NaOH + Al + H2O -> NaAlO2 + 3/2 H2

b) nH2= 0,6(mol)

-> nAl=0,4(mol) => mAl=0,4.27=10,8(g)

c) nAl=0,18((mol); nNaOH=0,2(mol)

PTHH: 0,18/1 < 0,2/1

=> Al hết, NaOH dư, tính theo nAl.

-> nH2= 3/2. 0,18=0,27(mol)

=>V(H2,đktc)=0,27.22,4= 6,048(l)

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

\(2NaOH+2Al+2H_2O\rightarrow2NaAlO_2+3H_2\)

\(...........0.4.........................0.6\)

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

\(n_{Al}=\dfrac{4.86}{27}=0.18\left(mol\right)\)

\(n_{NaOH}=\dfrac{8}{40}=0.2\left(mol\right)\)

\(2NaOH+2Al+2H_2O\rightarrow2NaAlO_2+3H_2\)

\(2.................2\)

\(0.2...............0.18\)

\(LTL:\dfrac{0.2}{2}>\dfrac{0.18}{2}\)

\(\Rightarrow NaOHdư\)

\(n_{H_2}=0.18\cdot\dfrac{3}{2}=0.27\left(mol\right)\)

\(V_{H_2}=0.27\cdot22.4=6.048\left(l\right)\)

a, PTHH: 2NaOH + 2Al + 2H₂O → 2NaAlO₂ + 3H₂↑

b, nH2=0,6 mol -mAl=10,8 g 

c, nAl=0,16(592) mol 

   nNaOH=0,2 mol 

-NaOH hết, Al dư 

-VH2=6,72 l

PTHH: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\uparrow\)

Ta có: \(\left\{{}\begin{matrix}n_{Zn}=\dfrac{26}{65}=0,4\left(mol\right)\\n_{H_2SO_4}=\dfrac{980\cdot10\%}{98}=1\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\) Axit còn dư, Kẽm p/ứ hết

\(\Rightarrow n_{H_2}=0,4\left(mol\right)\) \(\Rightarrow V_{H_2}=0,4\cdot22,4=8,96\left(l\right)\)

\(n_{Al}=\dfrac{2,5}{27}=\dfrac{25}{270}=\dfrac{5}{54}\left(mol\right)\\ n_{H_2SO_4}=0,5\left(mol\right)\\ a,2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ b,Vì:\dfrac{\dfrac{5}{54}}{2}< \dfrac{0,5}{4}\Rightarrow H_2SO_4dư\\ b,n_{H_2SO_4\left(dư\right)}=0,5-\dfrac{3}{2}.\dfrac{5}{54}=\dfrac{13}{36}\left(mol\right)\\ \Rightarrow m_{H_2SO_4}=\dfrac{13}{36}.98=\dfrac{637}{18}\left(g\right)\\ c,n_{H_2}=\dfrac{3}{2}.n_{Al}=\dfrac{3}{2}.\dfrac{5}{54}=\dfrac{5}{36}\left(mol\right)\\ \Rightarrow V_{H_2\left(đktc\right)}=\dfrac{5}{36}.22,4=\dfrac{28}{9}\left(l\right)\)

Tóm tắt:

m_Al= 2,5(g)

\(n_{H_2SO_4}=0,5\left(mol\right)\)

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a) PTHH?

b) Chất dư? m?

c) Thể tích khí sau phản ứng?

Anh ít tóm tắt lắm nên không biết tóm tắt như ri ổn không, vì em cần nên anh tóm tắt ra thử