a+b) PTHH: \(2Fe+3Cl_2\xrightarrow[]{t^o}2FeCl_3\)

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

Ta có: \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}n_{Cl_2}=0,3\left(mol\right)\\n_{HCl}=0,4\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Cl_2}=0,3\cdot71=21,3\left(g\right)\\V_{HCl}=\dfrac{0,4}{2}=0,2\left(l\right)=200\left(ml\right)\end{matrix}\right.\) 

Câu 1:

\(2Na+Br_2\rightarrow2NaBr\\ n_{NaBr}=\dfrac{61,8}{103}=0,6\left(mol\right)\\ n_{Na}=n_{NaBr}=0,6\left(mol\right)\\ n_{Br_2}=\dfrac{0,6}{2}=0,3\left(mol\right)\\ \Rightarrow a=m_{Na}=0,6.23=13,8\left(g\right)\\ m_{Br_2}=0,3.160=48\left(g\right)\\ m_{ddBr_2}=\dfrac{48}{5\%}=960\left(g\right)\)

Câu 2:

\(2Fe+3Cl_2\rightarrow\left(t^o\right)2FeCl_3\\ n_{FeCl_3}=\dfrac{40,625}{162,5}=0,25\left(mol\right)\\ n_{Fe}=n_{FeCl_3}=0,25\left(mol\right)\\ \Rightarrow m=m_{Fe}=0,25.56=14\left(g\right)\\ n_{Cl_2}=\dfrac{3}{2}.0,25=0,375\left(mol\right)\\ V_{Cl_2\left(đktc\right)}=0,375.22,4=8,4\left(l\right)\)

$2Fe + 3Cl_2 \xrightarrow{t^o} 2FeCl_3$
$n_{FeCl_3} = \dfrac{32,5}{162,5} = 0,2(mol)$

$n_{Cl_2} = \dfrac{3}{2}n_{FeCl_3} = 0,3(mol)$
$V_{Cl_2} = 0,3.22,4 = 6,72(lít)$

PTPU

2Fe+ 3Cl₂\(\xrightarrow[]{to}\) 2FeCl₃

áp dụng định luật bảo toàn khối lượng ta có:

m_FeCl3= m_Fe+ m_Cl2

\(\Rightarrow\) m_Cl2= m_FeCl3- m_Fe

= 32,5- 11,2= 21,3( g)

a, \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

\(n_{H_2}=\dfrac{7,437}{24,79}=0,3\left(mol\right)\)

Theo PT: \(n_{Fe}=n_{H_2}=0,3\left(mol\right)\Rightarrow m_{Fe}=0,3.56=16,8\left(g\right)\)

b, \(n_{H_2SO_4}=n_{H_2}=0,3\left(mol\right)\Rightarrow C\%_{H_2SO_4}=\dfrac{0,3.98}{120}.100\%=24,5\%\)

c, m _{dd sau pư} = 16,8 + 120 - 0,3.2 = 136,2 (g)

d, \(n_{FeSO_4}=n_{H_2}=0,3\left(mol\right)\)

\(\Rightarrow C\%_{FeSO_4}=\dfrac{0,3.152}{136,2}.100\%\approx33,48\%\)

Tìm số mol Fe tham gia phản ứng :

n Fe = n H  = 0,15 mol, suy ra m Fe  = 8,4 gam.

Ta có: \(n_{Fe}=\dfrac{28}{56}=0,5\left(mol\right)\)

a, PT: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

______0,5____0,5_____0,5_____0,5 (mol)

b, m_H2SO4 = 0,5.98 = 49 (g)

c, m_FeSO4 = 0,5.152 = 76 (g)

d, \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)

____0,5__0,5 (mol)

⇒ m_Cu = 0,5.64 = 32 (g)

Bạn tham khảo nhé!