PTHH: \(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)

Ta có: \(n_{CuO}=\dfrac{29,4}{80}=0,3675\left(mol\right)=n_{CuSO_4}=n_{H_2SO_4}\)

\(\Rightarrow\left\{{}\begin{matrix}m_{CuSO_4}=0,3675\cdot160=58,8\left(g\right)\\m_{H_2SO_4}=0,3675\cdot98=36,015\left(g\right)\\V_{H_2SO_4}=\dfrac{0,3675}{1}=0,3675\left(l\right)=367,5\left(ml\right)\end{matrix}\right.\)

$a)$

$Mg+H_2SO_4\to MgSO_4+H_2$

$b)$

$n_{Mg}=\frac{2,4}{24}=0,1(mol)$

Theo PT: $n_{MgSO_4}=n_{Mg}=0,1(mol)$

$\to m_{MgSO_4}=0,1.120=12(g)$

$c)$

$CuO+H_2\xrightarrow{t^o}Cu+H_2O$

Theo PT: $n_{Cu}=n_{H_2}=n_{Mg}=0,1(mol)$

$\to m_{Cu}=0,1.64=6,4(g)$

Cu(OH)₂ + H₂SO₄ \(\rightarrow\) CuSO₄ + 2H₂O

nCu(OH)₂ = \(\dfrac{29,4}{98}=0,3mol\)

Theo pt: nH₂SO₄ = nCu(OH)₂ = 0,3 mol

=> mH₂SO₄ = 0,3.98 = 29,4g

VH₂SO₄ = 0,3:1 = 0,3l

1a. PTHH: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\)

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

\(n_{H_2SO_4}=\dfrac{14,7}{1.2+32+16.4}=0,15\left(mol\right)\)

Do \(\dfrac{0,2}{1}>\dfrac{0,15}{1}\) => Fe dư, H₂SO₄ hết.

- Theo PTHH \(\Rightarrow n_{H_2SO_4}=n_{FeSO_4}=n_{H_2}=0,15\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}m_{FeSO_4}=0,15.152=22,8\left(g\right)\\V_{H_2}=0,15.22,4=3,36\left(l\right)\end{matrix}\right.\)

a)

$K_2O + H_2O \to 2KOH$
$n_{KOH} = 0,4.1 = 0,4(mol) \Rightarrow m_{K_2O} = \dfrac{1}{2}n_{KOH} = 0,2(mol)$
$m_{K_2O}= 0,2.94 = 18,8(gam)$
Suy ra: $a = 42,8 - 18,8 = 24(gam)$
b)

$n_{CuO} = \dfrac{24}{80} = 0,3(mol)$
$CuO + 2HCl \to CuCl_2 + H_2O$
$n_{HCl} = 2n_{CuO} = 0,6(mol)$
$m_{dd\ HCl} = \dfrac{0,6.36,5}{7,3\%} = 300(gam)$
$V_{dd\ HCl} = \dfrac{300}{1,15} = 260,87(ml)$

a) \(n_{H_2SO_4}=\dfrac{5,88}{98}=0,06\left(mol\right)\)

PTHH: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

            0,04<--0,06------->0,02---------->0,06

\(\Rightarrow\left\{{}\begin{matrix}a=m_{Al}=0,04.27=1,08\left(g\right)\\V=V_{H_2}=0,06.22,4=1,344\left(l\right)\end{matrix}\right.\)

b) 

Cách 1: \(m=m_{Al_2\left(SO_4\right)_3}=0,02.342=6,84\left(g\right)\)

Cách 2: \(m_{H_2}=0,06.2=0,12\left(g\right)\)

Áp dụng ĐLBTKL:

\(m_{Al}+m_{H_2SO_4}=m_{Al_2\left(SO_4\right)_3}+m_{H_2}\)

\(\Rightarrow m=m_{Al_2\left(SO_4\right)_3}=1,08+5,88-0,12=6,84\left(g\right)\)

c) \(n_{O_2}=\dfrac{1,344}{22,4}=0,06\left(mol\right)\)

PTHH: \(2H_2+O_2\xrightarrow[]{t^o}2H_2O\)

Xét tỉ lệ: \(\dfrac{0,06}{2}< \dfrac{0,06}{1}\Rightarrow\) O₂ dư, H₂ hết

Theo PTHH: \(n_{O_2\left(p\text{ư}\right)}=\dfrac{1}{2}.n_{H_2}=0,03\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{O_2\left(d\text{ư}\right)}=0,06-0,03=0,03\left(mol\right)\\m_{O_2\left(d\text{ư}\right)}=0,03.32=0,96\left(g\right)\\V_{O_2\left(d\text{ư}\right)}=0,03.22,4=0,672\left(l\right)\end{matrix}\right.\)

Theo PTHH: \(n_{H_2O}=n_{H_2}=0,06\left(mol\right)\)

\(\Rightarrow m_{H_2O}=0,06.18=1,08\left(g\right)\)

\(n_{FeSO_4}=\dfrac{m}{M}=\dfrac{22,8}{152}=0,15\left(mol\right)\\ PTHH:Fe+H_2SO_4->FeSO_4+H_2\)

tỉ lệ           1    :     1           :       1       :    1

n(mol)        0,15<-----------------0,15

\(m_{Fe}=n\cdot M=0,15\cdot56=8,4\left(g\right)\\ =>B\)

B

\(n_{H_2SO_4}=\dfrac{98.5\%}{98}=0,05\left(mol\right)\\ PTHH:CuO+H_2SO_4\rightarrow CuSO_4+H_2O\\ n_{CuSO_4}=n_{CuO}=n_{H_2SO_4}=0,05\left(mol\right)\\ a,m_{CuO}=0,05.80=4\left(g\right)\\ b,m_{CuSO_4}=0,05.160=8\left(g\right)\\ m_{ddCuSO_4}=98+4=102\left(g\right)\\ C\%_{ddCuSO_4}=\dfrac{8}{102}.100\approx7,843\%\)

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