\(n_{Fe}=\frac{5,6}{56}=0,1mol\)

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

\(\Rightarrow n_{FeCl_2}=n_{Fe}=0,1mol;n_{HCl}=2n_{Fe}=0,2mol\)

\(\Rightarrow V_{dd_{HCl}}=\frac{0,2}{0,5}=0,4l\)

\(\Rightarrow C_{M_{FeCl_2}}=\frac{0,1}{0,4}=0,25M\)

\(a)Mg+2HCl\rightarrow MgCl_2+H_2\)
\(b)n_{Mg}=\dfrac{3}{24}=0,125mol\\ n_{HCl}=0,1.1=0,1mol\\ \Rightarrow\dfrac{0,125}{1}>\dfrac{0,1}{2}\Rightarrow Mg.dư\\ n_{H_2}=n_{MgCl_2}=\dfrac{0,1}{2}=0,05mol\\ V_{H_2}=0,05.24,79=1,2395l\\ c)C_{M_{MgCl_2}}=\dfrac{0,05}{0,1}=0,5M\)

Đoạn xét tỉ lệ phải là \(\dfrac{0,125}{1}>\dfrac{0,1}{2}\) em nhé.

Gọi số mol HCl là x mol

Giả sử KOH hết 

⇒ mKCl = 74,5.0,1 = 7,45 gam > 6,525 

⇒ KOH dư, HCl hết.

HCl+KOH→KCl+H2O

x                  →x                     (mol)

(0,1 –  x).56 + x.(39 + 35,5) = 6,525

⇒ x = 0,05 mol ⇒ CM = 0,5M.

Phần c thì làm ntn ạ

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

\(2Al+6HCl->2AlCl_3+3H_2\) (1)

theo (1) \(n_{H_2}=\frac{3}{2}n_{Al}=0,6\left(mol\right)\) 

=> \(V_{H_2}=0,6.22,4=13,44\left(l\right)\)

Cho 10,08 g nhom tac dung vua du voi dung dich axit HCl.2M

a) viet phuong trinh phan ung va tinh the tich H2(dktc)

b) tinh the tich dung dich axit HCl.2M da dung

a,,mol HCl=CM\(\times\) V =0,5\(\times\)0,2=0,1   b,,: molHCL= 0,6.0,5=0,3mol

d, tổng thể tick sau trộn =200+600=800(ml)=0,8(l)   → molHCl sau trộn = 0,3+0,1=0,4mol

→Nồng độ sau HCl= \(\frac{n}{V}=\frac{0,4}{0,8}=0,5M\)

câu trả lời đúng là A

\(n_{Zn}=\dfrac{6.5}{65}=0.1\left(mol\right)\)

\(n_{H_2SO_4}=0.3\cdot1=0.3\left(mol\right)\)

\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

\(0.1........0.1...........0.1.......0.1\)

\(\Rightarrow H_2SO_4dư\)

\(n_{H_2SO_4\left(dư\right)}=0.3-0.1=0.2\left(mol\right)\)

\(n_{ZnSO_4}=n_{H_2}=0.1\left(mol\right)\)

\(C_{M_{ZnSO_4}}=\dfrac{0.1}{0.3}=0.33\left(M\right)\)

\(C_{M_{H_2SO_4\left(dư\right)}}=\dfrac{0.2}{0.3}=0.66\left(M\right)\)

\(a) Zn + H_2SO_4 \to ZnSO_4 + H_2\\ b) n_{Zn} = \dfrac{6,5}{65} = 0,1 < n_{H_2SO_4} =0,3 \to H_2SO_4\ dư\\ n_{H_2SO_4\ pư} = n_{ZnSO_4} = n_{Zn} = 0,1(mol)\\ n_{H_2SO_4\ dư} = 0,3 - 0,1 = 0,2(mol)\\ c) C_{M_{ZnSO_4}} = \dfrac{0,1}{0,3} = 0,33M\\ C_{M_{H_2SO_4}} = \dfrac{0,2}{0,3} = 0,67M\)

Bài 1:

n_Fe = \(\dfrac{5,6}{56}=0,1\left(mol\right)\)

n_HCl = 1 . 0,1 = 0,1 mol

Pt: Fe + 2HCl --> FeCl₂ + H₂

0,05<----0,1-----> 0,05---> 0,05....(mol)

Xét tỉ lệ mol giữa Fe và HCl:

\(\dfrac{0,1}{1}>\dfrac{0,1}{2}\)

Vậy Fe dư

V_H2 tạo ra = 0,05 . 22,4 = 1,12 (lít)

m_{Fe dư} = (0,1 - 0,05) . 56 = 2,8 (g)

C_{M FeCl2} = \(\dfrac{0,05}{0,1}=0,5M\)

Bài 2:

n_H2 = \(\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

Pt: Zn + 2HCl --> ZnCl₂ + H₂

0,1 mol<-0,2 mol<---------0,1 mol

a = 0,1 . 65 = 6,5 (g)

C_{M HCl} = \(\dfrac{0,2}{0,4}=0,5M\)

a, \(n_{HNO_3}=0,3.1=0,3\left(mol\right)\)

\(n_{Ba\left(OH\right)_2}=0,1.1=0,1\left(mol\right)\)

PT: \(2HNO_3+Ba\left(OH\right)_2\rightarrow Ba\left(NO_3\right)_2+2H_2O\)

Xét tỉ lệ: \(\dfrac{0,3}{2}>\dfrac{0,1}{1}\), ta được HNO₃ dư.

Theo PT: \(\left\{{}\begin{matrix}n_{Ba\left(NO_3\right)_2}=n_{Ba\left(OH\right)_2}=0,1\left(mol\right)\\n_{HNO_3\left(pư\right)}=2n_{Ba\left(OH\right)_2}=0,2\left(mol\right)\end{matrix}\right.\)

⇒ n_{HNO3 (dư)} = 0,3 - 0,2 = 0,1 (mol)

\(\Rightarrow\left\{{}\begin{matrix}C_{M_{Ba\left(NO_3\right)_2}}=\dfrac{0,1}{0,3+0,1}=0,25\left(M\right)\\C_{M_{HNO_3\left(dư\right)}}=\dfrac{0,1}{0,3+0,1}=0,25\left(M\right)\end{matrix}\right.\)

b, Ta có: \(n_{Na_2CO_3}=0,25.0,5=0,125\left(mol\right)\)

PT: \(Na_2CO_3+2HNO_3\rightarrow2NaNO_3+CO_2+H_2O\)

______0,05______0,1_______________0,05 (mol)

⇒ V_CO2 = 0,05.22,4 = 1,12 (l)

\(Na_2CO_3+Ba\left(NO_3\right)_2\rightarrow2NaNO_3+BaCO_{3\downarrow}\)

0,075________0,075_______________0,075 (mol)

⇒ m_BaCO3 = 0,075.197 = 14,775 (g)