Hóa học

\(n_{HCl}=\dfrac{200.14,6\%}{36,5}=0,8\left(mol\right)\)

\(CuO+2HCl\rightarrow CuCl_2+H_2O\)

  0,4         0,8                           ( mol )

\(m_{CuO}=0,4.80=32\left(g\right)\)

Đề bài hỏi gì và còn cho gì thêm không em?

\(a,Na_2CO_3+2HCl\rightarrow2NaCl+CO_2+H_2O\\ n_{CO_2}=\dfrac{0,448}{22,4}=0,02\left(mol\right)\\ n_{Na_2CO_3}=n_{CO_2}=0,02\left(mol\right)\\ n_{HCl}=0,02.2=0,04\left(mol\right)\\ C_{MddHCl}=\dfrac{0,04}{0,02}=2\left(M\right)\\ b,m_{NaCl\left(sau\right)}=m_{NaCl\left(bđ\right)}+m_{NaCl\left(sp\right)}=\left(5-0,02.106\right)+0,02.2.58,5=5,22\left(g\right)\\ c,\%m_{Na_2CO_3}=\dfrac{0,02.106}{5}.100\%=42,4\%\\ \Rightarrow\%m_{NaCl}=100\%-42,4\%=57,6\%\)

\(n_{CO_2}=\dfrac{0,448}{22,4}=0,02\left(mol\right)\)

\(Na_2CO_3+2HCl\rightarrow2NaCl+CO_2\uparrow+H_2O\) 

  0,02             0,04         0,02     0,02                ( mol )

\(C_{M_{HCl}}=\dfrac{0,04}{0,02}=2\left(M\right)\)

\(m_{Na_2CO_3}=0,02.106=2,12\left(g\right)\)

\(m_{NaCl\left(hh\right)}=5-2,12=2,88\left(g\right)\)

\(m_{muối}=2,88+\left(0,02.58,5\right)=4,05\left(g\right)\)

`c.\(\left\{{}\begin{matrix}\%m_{Na_2CO_3}=\dfrac{2,12}{5}.100=42,4\%\\\%m_{NaCl}=100\%-42,4\%=57,6\%\end{matrix}\right.\)

a. Vì NaCl ko phản ứng với HCl

\(n_{CO_2}=\dfrac{0.448}{22,4}=0,02\left(mol\right)\)

PTHH : Na₂CO₃ + 2HCl -> 2NaCl + CO₂ + H₂O

                 0,02        0,04       0,04      0,02

\(C_{M\left(HCl\right)}=\dfrac{0.04}{0,02}=2\left(M\right)\)

b. \(m_{NaCl}=0,04.58,5=2,34\left(g\right)\)

\(m_{Na_2CO_3}=0,02.106=2,12\left(g\right)\)

\(\Rightarrow\) m muối sau phản ứng là : \(\left(5-2,12\right)+2,34=5,22\left(g\right)\)

c. \(\%m_{Na_2CO_3}=\dfrac{2.12}{5}.100=42,4\%\)

\(\%m_{NaCl}=100\%-42,4\%=57,6\%\)

a)

$4P + 5O_2 \xrightarrow{t^o} 2P_2O_5$
b) $n_P = \dfrac{3,72}{31} = 0,12(mol) ; n_{O_2} = \dfrac{4,032}{22,4} = 0,18(mol)$

Ta thấy : 

$n_P : 4 < n_{O_2} : 5$ nên $O_2$ dư

$n_{P_2O_5} = \dfrac{1}{2}n_P = 0,06(mol)$
$m_{P_2O_5} = 0,06.142 = 8,52(gam)$

c) Hiện tượng : quỳ tím hoá đỏ

$P_2O_5 + 3H_2O \to 2H_3PO_4$

$n_{H_3PO_4} = 2n_{P_2O_5} = 0,12(mol)$
$C_{M_{H_3PO_4}} = \dfrac{0,12}{0,3} = 0,4M$

\(a,PTHH:4P+5O_2\rightarrow\left(t^o\right)2P_2O_5\\ b,n_P=\dfrac{3,72}{31}=0,12\left(mol\right)\\ n_{O_2}=\dfrac{4,032}{22,4}=0,18\left(mol\right)\\ Vì:\dfrac{0,12}{4}< \dfrac{0,18}{5}\Rightarrow O_2dư\\ n_{P_2O_5}=\dfrac{1}{2}.n_P=\dfrac{0,12}{2}=0,06\left(mol\right)\\ \Rightarrow m_{P_2O_5}=142.0,06=8,52\left(g\right)\\ c,P_2O_5+3H_2O\rightarrow2H_3PO_4\\ ddA:ddH_3PO_4\\ \Rightarrow Làm.quỳ.tím.hoá.đỏ\\ n_{H_3PO_4}=2.n_{P_2O_5}=2.0,06=0,12\left(mol\right)\\ C_{MddH_3PO_4}=\dfrac{0,12}{0,3}=0,4\left(M\right)\)

a. \(n_P=\dfrac{3.72}{31}=0,12\left(mol\right)\)

\(n_{O_2}=\dfrac{4.032}{22,4}=0,18\left(mol\right)\)

PTHH : 4P + 5O₂ --t^o---> 2P₂O₅

              0,12                      0,06

Ta thấy : \(\dfrac{0.12}{4}< \dfrac{0.18}{5}\) => P đủ , O₂ dư

b. \(m_{P_2O_5}=0,06.142=8,52\left(g\right)\)

c. PTHH : P₂O₅ + 3H₂O -> 2H₃PO₄

                  0,6                    0,12

Cho H₃PO₄ vào quỳ tìm sẽ làm cho quỳ tím hóa đỏ

\(C_M=\dfrac{0.12}{0,3}=0,4\left(M\right)\)

`Zn + 2HCl -> ZnCl_2 + H_2 \uparrow`

`0,2`   `0,4`                          `0,2`          `(mol)`

`n_[Zn]=13/65=0,2(mol)`

`a)V_[H_2]=0,2.22,4=4,48(l)`

`b)C_[M_[HCl]]=[0,4]/[0,15]=2,67(M)`

\(a,4P+5O_2\rightarrow\left(t^o\right)2P_2O_5\\ b,2KMnO_4\rightarrow\left(t^o,xt\right)K_2MnO_4+MnO_2+O_2\\ c,2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ d,2Na+2H_2O\rightarrow2NaOH+H_2\\ e,CaO+H_2O\rightarrow Ca\left(OH\right)_2\\ f,CuO+H_2\rightarrow\left(t^o\right)Cu+H_2O\\ g,Fe_2O_3+3H_2\rightarrow\left(t^o\right)2Fe+3H_2O\\ h,P_2O_5+3H_2O\rightarrow2H_3PO_4\)

`Zn + 2HCl -> ZnCl_2 + H_2 \uparrow`

`0,1`   `0,2`           `0,1`     `0,1 `           `(mol)`

`n_[Zn]=[6,5]/65=0,1(mol)`

`a)V_[H_2]=0,1.22,4=2,24(l)`

`b)V_[dd HCl]=[0,2]/[0,2]=1(l)`

`c)C_[M_[ZnCl_2]]=[0,1]/1=0,1(M)`

\(n_{Zn}=\dfrac{6.5}{65}=0,1mol\)

a. \(Zn+2HCl->ZnCl_2+H_2\)

    0,1                                     0,1 mol

\(\Rightarrow V_{H_2}=0,1.22,4=2,24\left(L\right)\)

b. \(n_{HCl}=2n_{Zn}=0,2mol\)

=> \(V_{ddHCl}=\dfrac{n}{C_M}=\dfrac{0.2}{0,2}=1\left(L\right)\)

c. \(n_{ZnCl_2}=n_{Zn}\)

\(C_{M\left(ZnCl_2\right)}=\dfrac{0.1}{1}=0,1\left(M\right)\)

\(SO_3+2NaOH->Na_2SO_4+H_2O\)

\(SO_3+2NaOH\rightarrow Na_2SO_4+H_2O\\ SO_3+NaOH\rightarrow NaHSO_4\)

`d_[A// H_2]=[\overline{M_A}]/2=17`

    `=>\overline{M_A}=34`

 `=>[n_[NO]]/[n_[NO_2]]=|[34-46]/[34-30]|=3`

     `=>\bb A`

`n_[KOH]=0,15.0,1=0,015(M)`

`m_[KOH]=0,015.56=0,84(g)`

\(m_{NaCl}=30\%.80=24g\)

\(n_{NaCl}=\dfrac{24}{58,5}=0,41mol\)