\(n_{Fe}=\dfrac{5.6}{56}=0.1\left(mol\right)\)

\(n_{HCl}=\dfrac{10.95}{36.5}=0.3\left(mol\right)\)

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

Lập tỉ lệ : 

\(\dfrac{0.2}{1}>\dfrac{0.3}{2}\Rightarrow Fedư\)

Khi đó : 

\(n_{FeCl_2}=n_{H_2}=\dfrac{1}{2}\cdot n_{HCl}=\dfrac{1}{2}\cdot0.3=0.15\left(mol\right)\)

\(m_{FeCl_2}=0.15\cdot127=19.05\left(g\right)\)

\(V_{H_2}=0.15\cdot22.4=3.36\left(l\right)\)

\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\\ n_{HCl}=\dfrac{10,95}{36,5}=0,3\left(mol\right)\\ PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\\ Vì:\dfrac{0,1}{1}< \dfrac{0,3}{2}\\ \Rightarrow HCldư\\ \Rightarrow n_{FeCl_2}=n_{Fe}=n_{H_2}=0,1\left(mol\right)\\ \Rightarrow m_{FeCl_2}=127.0,1=12,7\left(g\right)\\ V_{H_2\left(đktc\right)}=0,1.22,4=2,24\left(l\right)\)

Ta có:n_Fe= 5,6/56=0,1(mol)

n_HCl=10,95/36,5=0,3(mol)

PTHH: Fe + 2 HCl -> FeCl₂ + H₂

Ta có: 0,3/2 > 0,1/1

=> HCl dư, Fe hết, tính theo n_Fe

-> n_H2=n_FeCl2=n_Fe=0,1(mol)

=> V_H2(đktc)=0,1.22,4=2,24(l)

m_FeCl2=0,1.127=12,7(g)

Em đăng câu hỏi sang môn hóa nha em ơi

a) nFe= 5,6/56=0,1(mol)

nHCl=10,95/36,5=0,3(mol)

PTHH: Fe + 2 HCl -> FeCl2 + H2

Ta có: 0,3/2 > 0,1/1

=> HCl dư, Fe hết, tính theo nFe

-> nH2=nFeCl2=nFe=0,1(mol)

=> V(H2,đktc)=0,1.22,4=2,24(l)

mFeCl2=0,1.127=12,7(g)

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

a) Ta có: \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\) \(\Rightarrow n_{HCl}=0,2mol\)

\(\Rightarrow m_{ddHCl}=\dfrac{0,2\cdot36,5}{10,95\%}\approx66,67\left(g\right)\)

b) Theo PTHH: \(n_{ZnCl_2}=n_{H_2}=n_{Zn}=0,1mol\)

\(\Rightarrow\left\{{}\begin{matrix}m_{ZnCl_2}=0,1\cdot136=13,6\left(g\right)\\m_{H_2}=0,1\cdot2=0,2\left(g\right)\end{matrix}\right.\)

Mặt khác: \(m_{dd}=m_{Zn}+m_{ddHCl}-m_{H_2}=72,97\left(g\right)\)

\(\Rightarrow C\%_{ZnCl_2}=\dfrac{13,6}{72,97}\cdot100\%\approx18,64\%\)

Câu 3 : 

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

a) Pt : \(Mg+2HCl\rightarrow MgCl_2+H_2|\)

            1            2             1          1

          0,2         0,4          0,2         0,2

b) \(n_{HCl}=\dfrac{0,2.2}{1}=0,4\left(mol\right)\)

\(m_{HCl}=0,4.36,5=14,6\left(g\right)\)

\(m_{ddHCl}=\dfrac{14,6.100}{10,95}=133,3\left(g\right)\)

c) \(n_{H2}=\dfrac{0,4.1}{2}=0,2\left(mol\right)\)

\(V_{H2\left(dktc\right)}=0,2.22,4=4,48\left(l\right)\)

d) \(n_{MgCl2}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)

⇒ \(m_{MgCl2}=0,2.95=19\left(g\right)\)

\(m_{ddspu}=4,8+133,3-\left(0,2.2\right)=137,7\left(g\right)\)

\(C_{MgCl2}=\dfrac{19.100}{137,7}=13,8\)⁰/₀

 Chúc bạn học tốt

a) \(m_{HCl}=200.10,95\%=21,9\left(g\right)\)

b) \(CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\)

x_______2x________x____x(mol)

\(n_{HCl}=\dfrac{200.10,95\%}{36,5}=0,6\left(mol\right)\)

Dung dịch A phải có HCl dư mới có thể trung hòa được NaOH.

\(n_{NaOH}=0,05.2=0,1\left(mol\right)\)

\(NaOH+HCl\rightarrow NaCl+H_2O\)

y________y______y(mol)

Ta có hpt:

\(\left\{{}\begin{matrix}2x+y=0,6\\y=0,1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,25\\y=0,1\end{matrix}\right.\)

\(\Rightarrow a=m_{CaCO_3}=100x=100.0,25=25\left(g\right)\\ V=V_{CO_2\left(đktc\right)}=22,4x=22,4.0,25=5,6\left(l\right)\)

c) 

\(m_{ddA}=25+200-0,25.44=214\left(g\right)\\ C\%_{ddCaCl_2}=\dfrac{0,25.111}{214}.100\approx12,967\%\\ C\%_{ddHCl\left(dư\right)}=\dfrac{0,1.36,5}{214}.100\approx1,706\%\)