Ta co: \(\dfrac{0,2}{3}< \dfrac{0,3}{1}\Rightarrow\) FeCl₃ dư

3KOH + FeCl₃ \(\rightarrow\) 3KCl + Fe(OH)₃

de: 0,2 0,3

pu: 0,2 \(\dfrac{1}{15}\) 0,2 \(\dfrac{1}{15}\)

spu: 0 \(\dfrac{7}{30}\) 0,2 \(\dfrac{1}{15}\)

\(m_{FeCl_3\left(dư\right)}=\dfrac{7}{30}.162,5\approx37,92g\)

\(m_{KCl}=0,2.74,5=14,9g\)

\(m_{Fe\left(OH\right)_3}=\dfrac{1}{15}.107\approx7,13g\)

Có: 4.(3-1):2 = 6 điểm   

\(n_{HCl}=\dfrac{7,3}{36,5}=0,2\left(mol\right)\)

\(n_{H_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

Theo đề HCl dư nên tính theo H₂

Mg + 2HCl \(\rightarrow\) MgCl₂ + H₂

de: 0,2 0,05

pu: 0,05 0,1 0,05 0,05

spu: 0,05 0,1 0,05 0

a, \(m_{Mg}=0,05.24=1,2g\)

b, \(m_{HCl\left(dư\right)}=0,1.36,5=3,65g\)

229

tích nhé 

đổi 22,4 cm³ = 0,0224 dm³

3360 cm³ = 33,6 dm³

\(n_{CO_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

\(n_{N_2}=\dfrac{0,0224}{22,4}=0,001\left(mol\right)\)

\(n_{SO_3}=\dfrac{33,6}{22,4}=0,15\left(mol\right)\)

\(m_{CO_2}=0,2.44=8,8g\)

\(m_{N_2}=28.0,001=0,028g\)

\(m_{SO_3}=0,15.80=12g\)

\(m_{hh}=8,8+0,028+12=20,828g\)

hien tuong hoa hoc : 1 ; 3 ; 5 ; 6 ; 4

gọi x la so mol cua Fe

y la so mol cua FeO

\(n_{H_2SO_4}=0,3.1=0,3\left(mol\right)\)

Fe + H₂SO₄ \(\rightarrow\) FeSO₄ + H₂

de: x \(\rightarrow\) x \(\rightarrow\) x \(\rightarrow\) x

FeO + H₂SO₄ \(\rightarrow\) FeSO₄ + H₂

de: y \(\rightarrow\) y \(\rightarrow\) y \(\rightarrow\) y

Ta co: 56x + 72y = 18,4

x + y = 0,3

\(\Rightarrow\) x = 0,2 y = 0,1

a, \(m_{Fe}=56.0,2=11,2g\)

\(m_{FeO}=72.0,1=7,2g\)

\(\%m_{Fe}=\dfrac{11,2}{18,4}.100\%\approx60,87\%\)

\(\%m_{FeO}=100-60,87\approx39,13\%\)

b, \(m_{FeSO_4}=152.\left(0,1+0,2\right)=45,6g\)

\(m_{dd}=18,4+300.1,65-0,4=513g\)

\(C\%=\dfrac{45,6}{513}.100\%\approx8,89\%\)

2, \(n_{P_2O_5}=\dfrac{14,2}{142}=0,1\left(mol\right)\)

\(m_{H_2O}=1.185=185g\)

P₂O₅ + 3H₂O \(\rightarrow\) 2H₃PO₄

de: 0,1 \(\rightarrow\) 0,2

\(m_{H_3PO_4}=0,2.98=19,6g\)

\(m_{dd}=m_{P_2O_5}+m_{H_2O}=199,2g\)

a, \(C\%_{H_3PO_4}=\dfrac{19,6}{199,2}.100\%\approx9,84\%\)

\(m_{NaOH}=\dfrac{200.20}{100}=40g\)

\(n_{NaOH}=\dfrac{40}{40}=1\left(mol\right)\)

\(\dfrac{1}{3}>\dfrac{0,2}{1}\Rightarrow NaOH\) dư

3NaOH +H₃PO₄ \(\rightarrow\) Na₃PO₄ + 3H₂O

de: 1 0,2

pu: 0,6 0,2 0,2 0,6

spu: 0,4 0 0,2 0,6

\(m_{NaOH\left(dư\right)}=40.0,4=16g\)

\(m_{Na_3PO_4}=0,2.164=32,8g\)

\(m_{ddspu}=m_{NaOH}+m_{H_3PO_4}=59,6g\)

\(C\%_{NaOH}=\dfrac{16}{59,6}.100\%\approx26,85\%\)

\(C\%_{Na_3PO_4}=\dfrac{32,8}{59,6}.100\%\approx55,03\%\)

(n_{H_2SO_4}=dfrac{19,6}{98}=0,2left(mol ight))

2Al + 3H₂SO₄ ( ightarrow) Al₂(SO₄)₃ + 3H₂

de: (dfrac{2}{15}) (leftarrow) 0,2 ( ightarrow) 0,3

a, (m_{Al}=dfrac{2}{15}.27=3,6g)

b, (V_{H_2}=22,4.0,3=6,72l)

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

gọi x là so mol CaO tac dung vs HCl

y la so mol CaO tac dung vs HNO₃

CaO + 2HCl \(\rightarrow\) CaCl₂ + H₂O

de: x \(\rightarrow\) 2x \(\rightarrow\) x \(\rightarrow\) x

CaO + 2HNO₃ \(\rightarrow\) Ca(NO₃)₂ + H₂O

de: y \(\rightarrow\) 2y \(\rightarrow\) y \(\rightarrow\) y

Ta co: \(C_{M_{HCl}}=C_{M_{HNO_3}}=0,2\)

\(\Rightarrow\dfrac{2x}{V}=\dfrac{2y}{V}\Rightarrow2x=2y\Rightarrow x=y\)

\(\Rightarrow x=y=\dfrac{0,2}{2}=0,1\left(mol\right)\)

\(\Rightarrow V=1l\)

\(m_{CaCl_2}=0,1.111=11,1g\)

\(m_{Ca\left(NO_3\right)_2}=0,1.164=16,4g\)