a) \(n_{Na}=\dfrac{4,6}{23}=0,2\left(mol\right)\)

PTHH: 2Na + 2H₂O ---> 2NaOH + H₂ => ddA là NaOH

            0,2----------------->0,2------>0,1

b) \(V_{H_2}=0,1.22,4=2,24\left(l\right)\)

c) \(C_{M\left(NaOH\right)}=\dfrac{0,2}{0,4}=0,5M\)

\(n_{K2O}=\dfrac{9,4}{94}=0,1\left(mol\right)\)

Pt : \(K_2O+H_2O\rightarrow2KOH|\)

         1           1              2

        0,1                        0,2

a) \(n_{KOH}=\dfrac{0,1.2}{1}=0,2\left(mol\right)\)

200ml = 0,2l

\(C_{M_{ddKOH}}=\dfrac{0,2}{0,2}=1\left(M\right)\)

b) Pt : \(KOH+HCl\rightarrow KCl+H_2O|\)

              1            1           1            1

            0,1          0,1

\(n_{HCl}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)

\(m_{HCl}=0,1.36,5=3,65\left(g\right)\)

\(m_{ddHCl}=\dfrac{3,65.100}{10}=36,5\left(g\right)\)

c) \(CO_2+2KOH\rightarrow K_2CO_3+H_2O|\)

        1          2               1              1

      0,05      0,1

\(n_{CO2}=\dfrac{0,1.1}{2}=0,05\left(mol\right)\)

\(V_{CO2\left(dktc\right)}=0,05.22,4=1,12\left(l\right)\)

 Chúc bạn học tốt

a) \(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\)

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

           0,4-->0,6---------->0,2------->0,6

=> \(C_{M\left(dd.H_2SO_4\right)}=\dfrac{0,6}{0,15}=4M\)

b) V_H2 = 0,6.22,4 = 13,44 (l)

c) \(C_{M\left(Al_2\left(SO_4\right)_3\right)}=\dfrac{0,2}{0,15}=\dfrac{4}{3}M\)

\(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\\ pthh:2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\) 
           0,4     0,6                   0,2             0,6 
\(C_M_{H_2SO_4}=\dfrac{0,6}{0,15}=4M\\ V_{H_2}=0,622,4=13,44L\) 
\(C_M=\dfrac{0,2}{0,15}=1,3M\)

\(n_K=\dfrac{39}{39}=1\left(mol\right)\\ 2K+2H_2O\rightarrow2KOH+H_2\\ n_{H_2}=\dfrac{1}{2}=0,5\left(mol\right)\\ a,V_{H_2\left(đktc\right)}=0,5.22,4=11,2\left(l\right)\\ b,n_{KOH}=n_K=1\left(mol\right)\\ C_{MddKOH}=\dfrac{1}{0,2}=5\left(M\right)\\ c,2H_2+O_2\rightarrow\left(t^o\right)2H_2O\\ n_{O_2}=\dfrac{0,5}{2}=0,25\left(mol\right)\\ V_{O_2\left(đktc\right)}=0,25.22,4=5,6\left(l\right)\)

\(1,n_{Mg}=\dfrac{7,2}{24}=0,3\left(mol\right)\)

PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\)

           0,3---->0,6------------------>0,3

\(2,C_{M\left(HCl\right)}=\dfrac{0,6}{0,3}=2M\\ 3,V_{H_2}=0,3.22,4=6,72\left(l\right)\)

C2: 

PTHH:      2Al+6HCl →2AlCl₃ +3H₂

a)

Ta có: 

\(+n_{Al}=\dfrac{8,1}{27}=0,3\left(mol\right)\)

\(+n_{HCl}=\dfrac{21,9}{36,5}=0,6\left(mol\right)\)

Biện luận: 

\(\dfrac{0,3}{2}>\dfrac{0,6}{6}\)

⇒Al dư, HCl pư hết.

\(+n_{Al}\)_dư =0,3-0,2=0,1(mol

\(+m_{Al}\)_dư =0,1.27=2,7(gam)

b)

\(+n_{AlCl_3}=0,2\left(mol\right)\)

⇒\(m_{AlCl_3}=0,2.133,5=26,7\left(gam\right)\)

c) PTHH:  H₂+CuO→Cu+H₂O

\(+n_{CuO}=n_{H_2}=0,3\left(mol\right)\)

\(+m_{CuO}=0,3.80=24\left(gam\right)\)

Chúc bạn học tốt.

\(1.\)

\(n_{H_2}=\dfrac{5.6}{22.4}=0.25\left(mol\right)\)

\(2N+2nHCl\rightarrow2NCl_n+nH_2\)

\(\dfrac{0.5}{n}.....0.5...............0.25\)

\(M_N=\dfrac{16.25}{\dfrac{0.5}{n}}=32.5n\left(\dfrac{g}{mol}\right)\)

\(BL:n=2\Rightarrow N=65\)

\(Nlà:Zn\)

Không tính được thể tích vì thiếu nồng độ mol nhé.

\(2.\)

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

\(n_{HCl}=\dfrac{21.9}{36.5}=0.6\left(mol\right)\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

\(0.2........0.6..........0.2...........0.3\)

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

\(m_{AlCl_3}=0.2\cdot133.5=26.7\left(g\right)\)

\(CuO+H_2\underrightarrow{t^0}Cu+H_2O\)

\(0.3.....0.3\)

\(m_{CuO}=0.3\cdot80=24\left(g\right)\)

Câu 1:

a, Giả sử kim loại N có hóa trị n.

PT: \(2N+2nHCl\rightarrow2NCl_n+nH_2\)

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

Theo PT: \(n_N=\dfrac{2}{n}n_{H_2}=\dfrac{0,5}{n}\left(mol\right)\)

\(\Rightarrow M_N=\dfrac{16,25}{\dfrac{0,5}{n}}=32,5n\)

Với n = 1 ⇒ M_N = 32,5 (loại)

Với n = 2 ⇒ M_N = 65 (nhận)

Với n = 3 ⇒ M_N = 97,5 (loại)

Vậy, N là kẽm (Zn).

b, Theo PT: \(n_{HCl}=2n_{H_2}=0,5\left(mol\right)\)

Phần này đề bài có cho thiếu nồng độ mol của dd HCl không bạn nhỉ?

Câu 2:

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

Ta có: \(n_{Al}=\dfrac{8,1}{27}=0,3\left(mol\right)\)

\(n_{HCl}=\dfrac{21,9}{36,5}=0,6\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,3}{2}>\dfrac{0,6}{6}\), ta được Al dư.

Theo PT: \(n_{Al\left(pư\right)}=\dfrac{1}{3}n_{HCl}=0,2\left(mol\right)\)

\(\Rightarrow n_{Al\left(dư\right)}=0,1\left(mol\right)\Rightarrow m_{Al\left(dư\right)}=0,1.27=2,7\left(g\right)\)

b, Theo PT: \(\left\{{}\begin{matrix}n_{AlCl_3}=\dfrac{1}{3}n_{HCl}=0,2\left(mol\right)\\n_{H_2}=\dfrac{1}{2}n_{HCl}=0,3\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{AlCl_3}=0,2.133,5=26,7\left(g\right)\\m_{H_2}=0,3.2=0,6\left(g\right)\end{matrix}\right.\)

c, PT: \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)

Theo PT: \(n_{CuO}=n_{H_2}=0,3\left(mol\right)\)

\(\Rightarrow m_{CuO}=0,3.80=24\left(g\right)\)

Bạn tham khảo nhé!

a)

$BaO + H_2O \to Ba(OH)_2$
$n_{Ba(OH)_2} = n_{BaO} = \dfrac{30,6}{153} = 0,2(mol)$
$C_{M_{Ba(OH)_2}} = \dfrac{0,2}{0,2} = 1M$

b)

$Ba(OH)_2 + 2HCl \to BaCl_2 + 2H_2O$
$n_{HCl} = 2n_{Ba(OH)_2} = 0,4(mol)$
$V_{dd\ HCl} = \dfrac{0,4}{0,5} = 0,8(lít)$

PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)

Ta có: \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,2\left(mol\right)\\n_{FeCl_2}=0,1\left(mol\right)=n_{H_2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{H_2}=0,1\cdot22,4=2,24\left(l\right)\\m_{FeCl_2}=0,1\cdot127=12,7\left(g\right)\\C_{M_{FeCl_2}}=\dfrac{0,1}{0,1}=1\left(M\right)\\C_{M_{HCl}}=\dfrac{0,2}{0,1}=2\left(M\right)\end{matrix}\right.\)

 \(PTHH:4Al+6HCl\rightarrow2Al_2Cl_3+3H_2\uparrow\)

\(n_{Al}=\frac{3,78}{27}=0,14\left(mol\right)\)

\(\Rightarrow n_{H_2}=\frac{3}{4}n_{Al}=0,105\left(mol\right)\)

\(V_{H_2}=0,105.22,4=2,352\left(l\right)\)

\(n_{HCl}=\frac{3}{2}n_{Al}=\frac{3}{2}.0,14=0,21\left(mol\right)\)

\(C_{M_{ddHCl}}=\frac{0,21}{0,2}=1,05\left(M\right)\)

\(n_{Al_2Cl_3}=\frac{1}{2}n_{Al}=\frac{1}{2}.0,14=0,07\left(mol\right)\)

\(m_{Al_2Cl_3}=0,07.160,5=11,235\left(g\right)\)