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

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

 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.b.\(n_{Al}=\dfrac{5,4}{27}=0,2mol\)

\(n_{H_2SO_4}=\dfrac{39,2}{98}=0,4mol\)

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

Xét: \(\dfrac{0,2}{2}\) < \(\dfrac{0,4}{3}\)                                    ( mol )

        0,2       0,3                  0,1               0,3         ( mol )

\(m_{H_2SO_4\left(dư\right)}=\left(0,4-0,3\right).98=9,8g\)

\(m_{Al_2\left(SO_4\right)_3}=0,1.342=34,2g\)

c.\(2H_2+O_2\rightarrow\left(t^o\right)2H_2O\)   

   0,3     0,15                        ( mol )

\(V_{kk}=V_{O_2}.5=\left(0,15.22,4\right).5=16,8l\)

PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)

Ta có: \(\left\{{}\begin{matrix}n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\\n_{HCl}=0,25\left(mol\right)\end{matrix}\right.\)

Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,25}{2}\) \(\Rightarrow\) HCl còn dư, Kẽm p/ứ hết

\(\Rightarrow\left\{{}\begin{matrix}n_{H_2}=0,1\left(mol\right)\\n_{HCl\left(dư\right)}=0,05\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{H_2}=22,4\cdot0,1=2,24\left(l\right)\\m_{HCl\left(dư\right)}=0,05\cdot36,5=1,825\left(g\right)\end{matrix}\right.\)

zn+2Hcl→zncl2+h2

a)        nZn=6.5/65=0.1(mol)

         ta có nZn/1=0.1/1<nHcl/2=0.25/2=0.125

           →zn hết ,hcl dư

         theo pt:nH2=nZn=0.1 (mol)

     Vh2=0.1*22.4=2.24

b)    sau pư zn hết ,Hcl dư 

theo pt nHcl=2nZn=0.1*2=0.2( mol )

sô mol Hcl dư là:

0.25-0.2=0.05(mol)

mHcl dư là:0.05*36.5=1.825

xong

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

b) $n_{Al} = 0,45(mol) ; n_{H_2SO_4} =\dfrac{219}{980} (mol)$

Ta thấy : 

$n_{Al} : 2 > n_{H_2SO_4} : 3$ nên Al dư

Theo PTHH : 

$n_{Al\ pư} = \dfrac{2}{3}n_{H_2SO_4} = \dfrac{73}{490} (mol)$
$m_{Al\ dư} = 12,15 - \dfrac{73}{490}.27 = 8,127(gam)$

c) $n_{Al_2(SO_4)_3} = \dfrac{1}{3}n_{H_2SO_4} = \dfrac{73}{930}(mol)$
$m_{muối} = \dfrac{73}{930}.342 = 25,48(gam)$

d) $V_{H_2} = \dfrac{219}{980}.22,4 = 5(lít)$

Hình như đề sai

a,\(n_{Al}=\dfrac{12,15}{27}=0,45\left(mol\right)\)

  \(m_{H_2SO_4}=109,5.20\%=21,9\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{21,9}{98}=0,2235\left(mol\right)\)

\(a.2Al+6HCl\rightarrow2AlCl_3+3H_2\\ n_{Al}=\dfrac{12,15}{27}=0,45\left(mol\right)\\ n_{HCl}=\dfrac{20\%.109,5}{36,5}=0,6\left(mol\right)\\ b.Vì:\dfrac{0,45}{2}>\dfrac{0,6}{6}\\ \Rightarrow Aldư\\ n_{Al\left(dư\right)}=0,45-\dfrac{2}{6}.0,6=0,25\left(mol\right)\\ m_{Al\left(dư\right)}=0,25.27=6,75\left(g\right)\\ c.n_{AlCl_3}=\dfrac{2}{6}.0,6=0,2\left(mol\right)\\ m_{AlCl_3}=0,2.133,5=26,7\left(g\right)\\ d.n_{H_2}=\dfrac{3}{6}.0,6=0,3\left(mol\right)\\ V_{H_2\left(đktc\right)}=0,3.22,4=6,72\left(l\right)\)

\(a/ Mg+H_2SO_4 \to MgSO_4+H_2\\ b/ n_{Mg}=\frac{3,6}{24}=0,15(mol)\\ n_{H_2SO_4}=0,168(mol)\\ b/\\ Mg \text{ hết}; H_2SO_4 \text{ dư}\\ n_{H_2SO_4(dư)}=0,168-0,15=0,018(mol)\\ m_{H_2SO_4}=0,018.98=1,764(g)\\ c/\\ n_{MgSO_4}=0,15(mol)\\ m_{MgSO_4}=0,15.120=18(g)\\ d/\\ n_{H_2}=0,15\\ V=0,15.22,4=3,36(l)\)

a, PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

Ta có: \(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,1}{2}>\dfrac{0,25}{6}\), ta được Al dư.

Theo PT: \(n_{H_2}=\dfrac{1}{2}n_{HCl}=0,125\left(mol\right)\Rightarrow V_{H_2}=0,125.22,4=2,8\left(l\right)\)

b, Theo PT: \(n_{Al\left(pư\right)}=\dfrac{1}{3}n_{HCl}=\dfrac{1}{12}\left(mol\right)\Rightarrow n_{Al\left(dư\right)}=0,1-\dfrac{1}{12}=\dfrac{1}{60}\left(mol\right)\)

\(\Rightarrow m_{Al}=\dfrac{1}{60}.27=0,45\left(g\right)\)