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

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

Theo PT: \(\left\{{}\begin{matrix}n_{HCl}=3n_{Al}=0,6\left(mol\right)\\n_{H_2}=\dfrac{3}{2}n_{Al}=0,3\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow m_{HCl}=0,6.36,5=21,9\left(g\right)\)

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

c, Cách 1:

Theo PT: \(n_{AlCl_3}=n_{Al}=0,2\left(mol\right)\)

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

Cách 2:

Ta có: \(m_{H_2}=0,3.2=0,6\left(g\right)\)

Theo ĐLBT KL, có: m_Al + m_HCl = m_AlCl3 + m_H2

⇒ m_AlCl3 = m_Al + m_HCl - m_H2 = 5,4 + 21,9 - 0,6 = 26,7 (g)

Bạn tham khảo nhé!

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

Ta có: \(\left\{{}\begin{matrix}n_{Al}=\dfrac{4,05}{27}=0,15\left(mol\right)\\n_{H_2SO_4}=\dfrac{294\cdot10\%}{98}=0,3\left(mol\right)\end{matrix}\right.\)

Xét tỉ lệ: \(\dfrac{0,15}{2}< \dfrac{0,3}{3}\) \(\Rightarrow\) Axit còn dư

\(\Rightarrow\left\{{}\begin{matrix}n_{Al_2\left(SO_4\right)_3}=0,075\left(mol\right)=n_{H_2SO_4\left(dư\right)}\\n_{H_2}=0,225\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{H_2}=0,225\cdot22,4=5,04\left(l\right)\\m_{Al_2\left(SO_4\right)_3}=0,075\cdot342=25,65\left(g\right)\\m_{H_2SO_4\left(dư\right)}=0,075\cdot98=7,35\left(g\right)\\m_{H_2}=0,225\cdot2=0,45\left(g\right)\end{matrix}\right.\)

Mặt khác: \(m_{dd}=m_{Al}+m_{ddH_2SO_4}-m_{H_2}=297,6\left(g\right)\)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{Al_2\left(SO_4\right)_3}=\dfrac{25,65}{297,6}\cdot100\%\approx8,62\%\\C\%_{H_2SO_4\left(dư\right)}=\dfrac{7,35}{297,6}\cdot100\%\approx4,47\%\end{matrix}\right.\)

$PTHH:Zn+2HCl\to ZnCl_2+H_2\uparrow$

$n_{Zn}=\dfrac{13}{65}=0,2(mol)$

Theo PT: $n_{ZnCl_2}=n_{H_2}=0,2(mol);n_{HCl}=0,4(mol)$

$a)m_{axit}=m_{HCl}=n.M=0,4.36,5=14,6(g)$

$b)m_{ZnCl_2}=n.M=0,2.136=27,2(g)$

$c)V_{H_2(đktc)}=n.22,4=0,2.22,4=4,48(lít)$

Số mol kẽm là : 

\(n=\dfrac{m}{M}=\dfrac{13}{65}=0,2\left(mol\right)\)

PTHH : Zn + 2HCL -> ZnCl₂ + H₂

1 2 1 1 

0,2 mol -> 0,4 mol 0,2 mol 0,2 mol 

a, Khối lượng HCL là : 

\(m=n.M=0,4.35,5=14,2\left(g\right)\)

b, Khối lượng ZnCL₂ là : 

\(m=n.M=0,1.136=13,6\left(g\right)\)

c, Thể tích H₂ là : V = n . 22,4 = \(0,1.22,4=2,24\left(l\right)\)

\(a,n_{Zn}=\dfrac{0,65}{65}=0,01\left(mol\right)\)

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

          0,01--->0,02---->0,01---->0,01

\(V_{H_2}=0,01.22,4=0,224\left(l\right)\\ b,m_{ZnCl_2}=0,01.136=1,36\left(g\right)\\ V_{ddHCl}=\dfrac{0,02}{2}=0,01\left(l\right)\)

a)

$Mg + 2HCl \to MgCl_2 + H_2$

Theo PTHH : 

$n_{MgCl_2} = n_{H_2} = \dfrac{5,6}{22,4} = 0,25(mol)$
$m_{MgCl_2} = 0,25.95 = 23,75(gam)$

b)

$n_{HCl} = 2n_{H_2} = 0,5(mol)$
$m_{dd\ HCl} = \dfrac{0,5.36,5}{7,3\%} = 250(gam)$

c)

$2K + 2H_2O \to 2KOH + H_2$
$n_K = 2n_{H_2} = 0,5(mol)$
$m_K = 0,5.39 = 19,5(gam)$

a) Zn + 2HCl →ZnCl₂  + H₂

b) nZn = 6,5/65 = 0,1 mol . Theo tỉ lệ pư => nH₂ = nZn = nZnCl₂ =0,1 mol <=> V_H2(đktc) = 0,1.22,4 = 2,24 lít.

c) mZnCl₂ = 0,1 . 136 = 13,6 gam

d) nHCl =2nZn = 0,2 mol => mHCl = 0,2.36,5= 7,3 gam

Cách 2: áp dụng định luật BTKL => mHCl = mZnCl₂ + mH₂ - mZn 

<=> mHCl = 13,6 + 0,1.2 - 6,5 = 7,3 gam

\(n_{Fe}=\dfrac{m}{M}=\dfrac{28}{56}=0,5\left(mol\right)\\ PTHH:Fe+2HCl->FeCl_2+H_2\)

ti le        1      :    2       :      1      :    1

n(mol)     0,5-->1--------->0,5------>0,5

\(m_{FeCl_2}=n\cdot M=0,5\cdot\left(56+35,5\cdot2\right)=63,5\left(g\right)\\ V_{H_2\left(dktc\right)}=n\cdot22,4=0,5\cdot22,4=11,2\left(l\right)\)

a)\(m_{ddHCl}=\dfrac{4,8}{10\%}.100\%=48\left(g\right)\)

b)\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)

\(PTHH:Mg+2HCl\xrightarrow[]{}MgCl_2+H_2\)

\(n_{MgCl_2}=n_{Mg}=n_{H_2}=0,2\)

\(m_{MgCl_2}=0,2.95=19\left(g\right)\)

\(V_{H_2}=0,2.22,4=4.48\left(l\right)\)

c)\(m_{H_2}=0,2.2=0,4\left(g\right)\)

\(m_{ddMgCl_2}=4,8+48-0,4=52,4\left(g\right)\)

\(C\%_{MgCl_2}=\dfrac{19}{52,4}.100\%=36\%\)

\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\\ n_{HCl}=\dfrac{109,5.20\%}{36,5}=0,6\left(mol\right)\\ PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ Vì:\dfrac{0,2}{1}< \dfrac{0,6}{2}\Rightarrow HCldư\\ n_{H_2}=n_{ZnCl_2}=n_{Zn}=0,2\left(mol\right)\\ n_{HCl\left(dư\right)}=0,6-0,2.2=0,2\left(mol\right)\\ a,V_{H_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\\ b,m_{ZnCl_2}=136.0,2=27,2\left(g\right)\\ c,m_{ddsau}=13+109,5-0,2.2=122,1\left(g\right)\\ C\%_{ddZnCl_2}=\dfrac{27,2}{122,1}.100\approx22,277\%\\ C\%_{ddHCl\left(dư\right)}=\dfrac{0,2.36,5}{122,1}.100\approx5,979\%\)

Zn + 2HCl -> ZnCl₂ + H₂ 

a, n_Zn = 13/65= 0,2(mol)

m_HCl= 109,5.20%/100%=21.9(g)

n_HCl=21,9/36,5=0,6(mol)

Theo PT n_HCl = 2n_Zn= 2.0,2= 0,4(mol)<0,6(mol)

=> HCl pư dư, Zn pư hết

Theo PT: n_H2= n_Zn =0,2(mol)

V_H2=0,2.22,4=4,48(l)

b, Theo PT: n_ZnCl2=n_Zn=0,2(mol)

m_ZnCl2= 0,2.136=27,2(g)

c, mdd sau pư= 13+109,5-0,2.2=122,1(g)

C%dd ZnCl2=27,2.100%/122,1=22,28%

n_HCl dư= 0,6-0,4=0,2(mol)

mHcl