a) PTHH: FeCl₃ + 3KOH → Fe(OH)₃ + 3KCl

b) Theo ĐLBTKL ta có:

 \(m_{FeCl_3}+m_{KOH}=m_{Fe\left(OH\right)_3}+m_{KCl}\)

\(\Leftrightarrow m_{FeCl_3}=m_{Fe\left(OH\right)_3}+m_{KCl}-m_{KOH}=2,14+4,47-3,36=3,25\left(g\right)\)

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

\(Fe+\dfrac{3}{2}Cl_2\underrightarrow{^{^{t^0}}}FeCl_3\)

\(0.1.......0.15..........0.1\)

\(V_{Cl_2}=0.15\cdot22.4=3.36\left(l\right)\)

\(m_{FeCl_3}=0.1\cdot162.5=16.25\left(g\right)\)

a.3.36l

b.13.25g

câu b là 16,25g nha bạn oi mik ghi nhầm

Sơ đồ

Sắt (III) sunfat + Natri hidroxit → Sắt (III) hidroxit + natri sunfat

Áp dụng ĐLBTKL, ta có

\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\\ 2Fe+3Cl_2\rightarrow\left(t^o\right)2FeCl_3\\ n_{Cl_2}=\dfrac{3}{2}.0,2=0,3\left(mol\right)\\ n_{FeCl_3}=n_{Fe}=0,2\left(mol\right)\\ a,V_{Cl_2\left(đktc\right)}=0,3.22,4=6,72\left(l\right)\\ b,m_{FeCl_3}=162,5.0,2=32,5\left(g\right)\)

a, \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)

PT: \(2Fe+3Cl_2\underrightarrow{t^o}2FeCl_3\)

Theo PT: \(n_{Cl_2}=\dfrac{3}{2}n_{Fe}=0,3\left(mol\right)\)

\(\Rightarrow V_{Cl_2}=0,3.22,4=6,72\left(l\right)\)

b, \(n_{FeCl_3}=n_{Fe}=0,2\left(mol\right)\Rightarrow m_{FeCl_3}=0,2.162,5=32,5\left(g\right)\)

Theo ĐLBTKL: 

\(m_{Fe_2\left(SO_4\right)_3}+m_{NaOH}=m_{Fe\left(OH\right)_3}+m_{Na_2SO_4}\)

=> \(m_{NaOH}=10,7+21,3-20=12\left(g\right)=>n_{NaOH}=\dfrac{12}{40}=0,3\left(mol\right)\)

=> D

\(n_{KOH}=\dfrac{100.14}{100.56}=0,25(mol)\\ 2KOH+CuCl_2\to Cu(OH)_2\downarrow+2KCl\\ \Rightarrow n_{CuCl_2}=n_{Cu(OH)_2}=0,125(mol);n_{KCl}=0,25(mol)\\ a,m_{CuCl_2}=0,125.135=16,875(g)\\ b,m_{Cu(OH)_2}=0,125.98=12,25(g)\\ c,C\%_{KCl}=\dfrac{0,25.74,5}{100+16,875-12,25}.100\%=17,8\%\\ d,Cu(OH)_2\xrightarrow{t^o}CuO+H_2O\\ \Rightarrow n_{CuO}=0,125(mol)\\ \Rightarrow m_{CuO}=0,125.80=10(g)\)

a) 

$Zn + 2HCl \to ZnCl_2 + H_2$
$Fe_2O_3 + 6HCl \to 2FeCl_3 + 3H_2O$

Theo PTHH : $n_{Zn} = n_{H_2} = 0,06(mol)$
$\Rightarrow n_{Fe_2O_3} = \dfrac{7,1-0,06.65}{160} = 0,02(mol)$

Theo PTHH : $n_{HCl} = 2n_{Zn} + 6n_{Fe_2O_3} = 0,24(mol)$
$m_{HCl} = 0,24.36,5 = 8,76(gam)$

b)

Gọi $n_{CuO} = x(mol) ; n_{Fe_3O_4} = y(mol) \Rightarrow 80a + 232y = 3,92(1)$
$CuO + H_2 \xrightarrow{t^o} Cu + H_2O$
$Fe_3O_4 + 4H_2 \xrightarrow{t^o} 3Fe + 4H_2O$

Theo PTHH : $n_{H_2} =x + 4y = 0,06(2)$

Từ (1)(2) suy ra:  x = 0,02;  y = 0,01

$n_{Cu} = 0,02(mol) \Rightarrow m_{Cu} = 0,02.64 = 1,28(gam)$
$n_{Fe} = 0,01.3 = 0,03(mol) \Rightarrow m_{Fe} = 0,03.56 = 1,68(gam)$

BT1 : 

Bảo toàn khối lượng : 

\(m_{FeCl_3}=m_{Fe}+m_{Cl_2}=11.2+21.3=32.5\left(g\right)\)

\(Fe+\dfrac{3}{2}Cl_2\underrightarrow{^{^{t^0}}}FeCl_3\)

\(1.5.....2.25......1.5\)

\(n_{Fe}=\dfrac{9\cdot10^{23}}{6\cdot10^{23}}=1.5\left(mol\right)\)

Số phân tử Cl₂ : \(2.25\cdot6\cdot10^{23}=13.5\cdot10^{23}\left(pt\right)\)

Số phân tử FeCl₃ : \(1.5\cdot6\cdot10^{23}=9\cdot10^{23}\left(pt\right)\)

BT2: 

\(C+O_2\underrightarrow{^{^{t^0}}}CO_2\)

Bảo toàn khối lượng : 

\(m_C+m_{O_2}=m_{CO_2}\)

\(m_C=m_{CO_2}-m_{O_2}=4.4-3.2=1.2\left(kg\right)\)

\(\%C=\dfrac{1.2}{1.5}\cdot100\%=80\%\)