\(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\\ n_{HCl}=\dfrac{10,95}{36,5}=0,3\left(mol\right)\\ pthh:Mg+2HCl\rightarrow MgCl_2+H_2\\ LTL:\dfrac{0,1}{1}< \dfrac{0,3}{2}\) 
=> HCl dư 

Pt: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

\(n_{Zn}=\dfrac{3,25}{65}=0,05mol\)

\(n_{HCl}=\dfrac{43,8}{36,5}=1,2mol\)

Lập tỉ lệ:

nZn : nHCl = \(\dfrac{0,05}{1}:\dfrac{1,2}{2}=0,05:0,6\)

=> HCl dư

nHCl dư = 1,2 - 0,05 = 1,15 mol

mHCl dư = 1,15 . 36,5 = 41,975g

b) Theo pt: nH₂ = nZn = 0,05 mol

=> VH₂ = 0,05.22,4=1,12lit

đề có sai chỗ 43,8 g HCl ko, số hơi to

\(a) n_{Zn} = \dfrac{3,25}{65} = 0,05(mol)\ ; n_{HCl} = \dfrac{43,8}{36,5} = 1,2(mol)\\ Zn + 2HCl \to ZnCl_2 + H_2\\ n_{HCl} = 1,2 > 2n_{Zn} = 0,1 \to HCl\ dư\\ n_{HCl\ pư} = 2n_{Zn} = 0,1(mol)\\ \Rightarrow m_{HCl\ dư} = 43,8 - 0,1.36,5 = 40,15(gam)\\ c) n_{H_2} = n_{Zn} = 0,05(mol) \Rightarrow V_{H_2} = 0,05.22,4 = 1,12(lít)\)

n Zn = 26/65 = 0,4(mol)

$Zn + 2HCl \to ZnCl_2 + H_2$
Ta thấy: 

 n Zn / 1 = 0,4 > n HCl / 2 = 0,25 nên Zn dư

Theo PTHH :

n Zn pư = 1/2 n HCl = 0,25(mol)

Suy ra:  m Zn dư = 26 - 0,25.65 = 9,75 gam

\(n_{Zn}=\dfrac{26}{65}=0.4\left(mol\right)\)

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

\(1........2\)

\(0.4........0.5\)

\(LTL:\dfrac{0.4}{1}>\dfrac{0.5}{2}\Rightarrow Zndư\)

\(m_{Zn\left(dư\right)}=\left(0.4-0.25\right)\cdot65=9.75\left(g\right)\)

\(n_{Zn}=\dfrac{13}{65}=0.2\left(mol\right)\)

\(n_{HCl}=\dfrac{182.5\cdot10}{100\cdot36.5}=0.5\left(mol\right)\)

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

\(0.2......0.4..........0.2........0.2\)

\(n_{HCl\left(dư\right)}=0.5-0.4=0.1\left(mol\right)\)

\(m_{HCl\left(dư\right)}=0.1\cdot36.5=3.65\left(g\right)\)

\(V_{H_2}=0.2\cdot22.4=4.48\left(l\right)\)

\(m_{\text{dung dịch sau phản ứng}}=13+182.5-0.2\cdot2=195.1\left(g\right)\)

\(C\%_{HCl\left(dư\right)}=\dfrac{3.65}{195.1}\cdot100\%=1.87\%\)

\(C\%_{ZnCl_2}=\dfrac{0.2\cdot136}{195.1}\cdot100\%=13.94\%\)

Bài 2:

\(a.n_{CaCl_2}=\dfrac{22,2}{111}=0,2\left(mol\right)\\ n_{AgNO_3}=\dfrac{1,7}{170}=0,01\left(mol\right)\\ CaCl_2+2AgNO_3\rightarrow Ca\left(NO_3\right)_2+2AgCl\\ a.Vì:\dfrac{0,2}{1}>\dfrac{0,01}{2}\Rightarrow CaCl_2dư\\b.n_{AgCl}=n_{AgNO_3}=0,01\left(mol\right)\\ \Rightarrow m_{\downarrow}=m_{AgCl}=143,5.0,01=1,435\left(g\right)\)

Bài 1:

\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\\ n_{HCl}=0,1.3=0,3\left(mol\right)\\ Zn+2HCl\rightarrow ZnCl_2+H_2\\ a,Vì:\dfrac{0,3}{2}< \dfrac{0,2}{1}\Rightarrow Zndư\\ b.n_{ZnCl_2}=\dfrac{0,3}{2}=0,15\left(mol\right)\\ m_{ZnCl_2}=136.0,15=20,4\left(g\right)\)

1.

a, \(n_{Mg}=\dfrac{3,6}{24}=0,15\left(mol\right);n_{HCl}=\dfrac{18,25}{36,5}=0,5\left(mol\right)\)

PTHH: Mg + 2HCl → MgCl₂ + H₂

Mol:     0,15     0,3

b, Ta có: \(\dfrac{0,15}{1}< \dfrac{0,5}{2}\) ⇒ Mg pứ hết, HCl dư

\(m_{HCldư}=\left(0,5-0,3\right).36,5=7,3\left(g\right)\)

c, \(V_{H_2}=0,15.22,4=3,36\left(l\right)\)

2.

a, \(n_P=\dfrac{6,2}{31}=0,2\left(mol\right)\)

PTHH: 4P + 5O₂ ---t^o→ 2P₂O₅

Mol:     0,2    0,25              0,1

b, \(V_{O_2}=0,25.22,4=5,6\left(l\right)\)

c, \(m_{P_2O_5}=0,1.142=14,2\left(g\right)\)

d, \(V_{kk}=5,6.5=28\left(l\right)\)

- Hỗn hợp rắn sau pư gồm: Fe và FeCl₃.

Ta có: \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

BT e, có: \(2n_{Fe}=2n_{H_2}\Rightarrow n_{Fe}=0,1\left(mol\right)\)

BTNT Fe, có: \(n_{FeCl_3}=n_{Fe\left(OH\right)_3}=0,03\left(mol\right)\) = n_{Fe (pư)}

BTNT Fe, có: n_{Fe (ban đầu)} = n_Fe + n_FeCl3 = 0,13 (mol)

\(\Rightarrow H\%=\dfrac{0,03}{0,13}.100\%\approx23,077\%\)