Phương trình hoá học : 2Fe + 3 Cl 2 → t ° FeCl 3

Theo định luật bảo toàn khối lượng :

m Fe + m Cl 2 = m FeCl 3

  m Cl 2 = m FeCl 3 - m Fe = 16,25 - 5,6 = 10,65g

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

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

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

Chọn đáp án A

n F e C l 3 = 32 , 5 162 , 5 = 0 , 2 m o l

Bảo toàn nguyên tố Cl

2 n C l 2 = 3 n F e C l 3 ⇒ n C l 2 = 0 , 3 m o l ⇒ V = 0 , 3.22 , 4 = 6 , 72 l

\(a.Sắt+Clo\rightarrow Sắt\left(III\right)clorua\\ b.2Fe+3Cl_2-^{t^o}\rightarrow2FeCl_3\\ c.m_{Fe}+m_{Cl_2}=m_{FeCl_3}\\ \Rightarrow m_{FeCl_3}=5,6+10,65=16,25\left(g\right)\)

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+3Cl₂-to>2FeCl₃

_{0,25------0,375------0,25 mol}

n _FeCl3=\(\dfrac{40,625}{162,5}\)=0,25 mol

=>m _Fe=0,25.56=14g

=>V _Cl2=0,375.22,4=8,4g

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

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

Câu 1:

\(Mg+Br_2\rightarrow MgBr_2\\ n_{Br_2}=\dfrac{11,2}{160}=0,07\left(mol\right)=n_{Mg}=n_{MgBr_2}\\ a=m_{Mg}=0,07.24=1,68\left(g\right)\\ m_{MgBr_2}=184.0,07=12,88\left(g\right)\)

Mg+Br2->MgBr2

0,07--0,07----0,07

n Br2=\(\dfrac{11,2}{160}\)=0,07 mol

=>m Mg=0,07.24=1,68g

=>m MgBr2=0,07.184=12,88g

Câu 2:

\(n_{Al}=\dfrac{14,58}{27}=0,54\left(mol\right)\\ 2Al+3Cl_2\rightarrow\left(t^o\right)2AlCl_3\\ n_{AlCl_3}=n_{Al}=0,54\left(mol\right)\\ n_{Cl_2}=\dfrac{3}{2}.0,54=0,81\left(mol\right)\\ V_{Cl_2\left(đktc\right)}=0,81.22,4=18,144\left(l\right)\\ m_{AlCl_3}=0,54.133,5=72,09\left(g\right)\)

\(n_{Fe}=\dfrac{0,56}{56}=0,01\left(mol\right)\)

PTHH: 2Fe + 3Cl₂ --t^o--> 2FeCl₃

          0,01-->0,015----->0,01

=> V = 0,015.22,4 = 0,336 (l)

m_FeCl3 = 0,01.162,5 = 1,625 (g)