\(\%Ca=\dfrac{40.1}{56}.100\%=71,4\%\\ \%O=\dfrac{16.1}{56}.100\%=28,6\%\)

a) PTHH: C + O2 -to-> CO2

x_____________x_____x(mol)

S+ O2 -to-> SO2

y__y________y(mol)

b) Ta có: 

\(\left\{{}\begin{matrix}12x+32y=5,6\\32x+32y=9,6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,2\\y=0,1\end{matrix}\right.\)

mC=0,2.12=2,4(g)

mS=0,1.32=3,2(g)

c) 

\(\%mC=\dfrac{2,4}{5,6}.100\approx42,857\%\\ \rightarrow\%mS\approx100\%-42,857\%\approx57,143\%\)

d)

 \(\%nCO2=\dfrac{x}{x+y}.100\%=\dfrac{0,2}{0,2+0,1}.100\approx66,667\%\\ \rightarrow\%nSO2=\dfrac{y}{x+y}.100\%=\dfrac{0,1}{0,2+0,1}.100\approx33,333\%\)

a)nO2=m/M=9,6/32=0,3 (mol)

C + O2 ->t° CO2

1:1:1

x/12 :(x/12) :x/12 mol

S + O2->t° SO2

1:1:1

5,6-x/32: (5,6-x/32): 5,6-x/32 mol

gọi x là số gam của cacbon 

nC=m/M=x/12(mol)

nS=5,6-x/12 (mol)

b)ta có phương trinh

5,6-x/32+x/12=0,3

<=>3(5,6-x)/96 + 8x/96= 28,8/96

->3(5,6-x)+8x=28,8

<=> 16,8 -3x+8x=28,8

<=>-3x+8x=12

<=>5x=12

<=>x=2,4

-> mC=2,4(g)

mS=5,6-2,4=3,2(g)

c)%mC=2,4/5,6.100%= 42,857%

%mS=100%-42,857%=57,143%

d)%nCO2=0,2/0,3.100%=66,7%

%nSO2=100%-66,7%=33,3%

a)

$Fe + 2HCl \to FeCl_2 + H_2$
$FeO +2 HCl \to FeCl_2 + H_2O$

b)

Theo PTHH : 

$n_{Fe} = n_{H_2} = \dfrac{4,48}{22,4} = 0,2(mol)$

$\%m_{Fe} = \dfrac{0,2.56}{20}.100\% = 56\%$

$\%m_{FeO} = 100\% - 56\% = 44\%$

c) $n_{FeO} = \dfrac{11}{90}(mol)$
$n_{HCl} = 2n_{Fe} + 2n_{FeO} = \dfrac{29}{45}(mol)$

$m_{dd\ HCl} = \dfrac{ \dfrac{29}{45}.36,5}{7,3\%} = 322,22(gam)$

a, PT: \(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)

\(FeO+H_2SO_4\rightarrow FeSO_4+H_2O\)

Gọi: \(\left\{{}\begin{matrix}n_{CuO}=x\left(mol\right)\\n_{FeO}=y\left(mol\right)\end{matrix}\right.\) ⇒ 80x + 72y = 11,2 (1)

Ta có: \(n_{H_2SO_4}=0,15.1=0,15\left(mol\right)\)

Theo PT: \(n_{H_2SO_4}=n_{CuO}+n_{FeO}=x+y=0,15\left(2\right)\)

Từ (1) và (2) ⇒ x = 0,05 (mol), y = 0,1 (mol)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{CuO}=\dfrac{0,05.80}{11,2}.100\%\approx35,71\%\\\%m_{FeO}\approx64,28\%\end{matrix}\right.\)

b, Theo PT: \(\left\{{}\begin{matrix}n_{CuSO_4}=n_{Cu}=0,05\left(mol\right)\\n_{FeSO_4}=n_{FeO}=0,1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}C_{M_{CuSO_4}}=\dfrac{0,05}{0,15}=\dfrac{1}{3}\left(M\right)\\C_{M_{FeSO_4}}=\dfrac{0,1}{0,15}=\dfrac{2}{3}\left(M\right)\end{matrix}\right.\)

a) Pt : \(Mg+2HCl\rightarrow MgCl_2+H_2|\)

             1         2                1          1

            0,3     0,6                            0,3

           \(MgO+2HCl\rightarrow MgCl_2+H_2O|\)

               1           2               1           1

              0,1        0,2

b) \(n_{Mg}=\dfrac{0,3.1}{1}=0,3\left(mol\right)\)

\(m_{Mg}=0,3.24=7,2\left(g\right)\)

\(m_{MgO}=11,2-7,2=4\left(g\right)\)

c) ⁰/₀Mg = \(\dfrac{7,2.100}{11,2}=64,29\)⁰/₀

     ⁰/₀MgO = \(\dfrac{4.100}{11,2}=35,71\)⁰/₀

d) Có : \(m_{MgO}=4\left(g\right)\)

\(n_{MgO}=\dfrac{4}{40}=0,1\left(mol\right)\)

\(n_{HCl\left(tổng\right)}=0,6+0,2=0,8\left(mol\right)\)

\(m_{HCl}=0,8.36,5=29,2\left(g\right)\)

\(C_{ddHCl}=\dfrac{29,2.100}{200}=14,6\)⁰/₀

 Chúc bạn học tốt

a)

$Fe + H_2SO_4 \to FeSO_4 + H_2$
$FeO + H_2SO_4 \to FeSO_4 + H_2O$
$FeSO_4 + 7H_2O \to FeSO_4.7H_2O$

b)

n Fe = a(mol) ; n FeO = b(mol)

=> 56a + 72b = 54,4(1)

Theo PTHH : 

n FeSO4.7H2O = a + b = 222,4/278 = 0,8(2)

Từ (1)(2) suy ra a = 0,2 ; b = 0,6

Suy ra : 

%m Fe = 0,2.56/54,4  .100% = 20,59%

%m FeO = 100% -20,59% = 79,41%

a, \(n_{H_2SO_4}=0,45.0,2=0,09\left(mol\right)\)

PTHH: FeO + H₂SO₄ → FeSO₄ + H₂O

Mol:       a          a

PTHH: MgO + H₂SO₄ → MgSO₄ + H₂O

Mol:       b              b

Ta có: \(\left\{{}\begin{matrix}72a+40b=4,48\\a+b=0,09\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,0275\\b=0,0625\end{matrix}\right.\)

\(\%m_{FeO}=\dfrac{0,0275.72.100\%}{4,48}=44,196\%\)

\(\%m_{MgO}=100-44,196=55,804\%\)

b, 

PTHH: FeO + H₂SO₄ → FeSO₄ + H₂O

Mol:   0,0275                   0,0275

PTHH: MgO + H₂SO₄ → MgSO₄ + H₂O

Mol:    0,0625                     0,0625

\(C_{M_{ddFeSO_4}}=\dfrac{0,0275}{0,2}=0,1375M\)

\(C_{M_{ddMgSO_4}}=\dfrac{0,0625}{0,2}=0,3125M\)