\(M_{hh}=2.17,2=34,4g/mol\)

Khối lượng hỗn hợp là: \(m_{1hh}=n_{hh}.M_{hh}=0,1.34,4=3,44g\)

Đặt \(\hept{\begin{cases}x\left(mol\right)=CO\\y\left(mol\right)=O_2\end{cases}}\)

Có \(\hept{\begin{cases}x+y=0,1\\28x+32y=3,44\end{cases}}\)

\(\rightarrow\hept{\begin{cases}x=-0,06\\y=0,16\end{cases}}\)

Bạn xem lại đề nhé

Đặt \(\hept{\begin{cases}a\left(mol\right)=n_{H_2}=n_{O_2\left(A\right)}\\2b\left(mol\right)=n_{Cl_2}\\3b\left(mol\right)=n_{O_2\left(B\right)}\end{cases}}\)

\(\overline{M_A}=\frac{2a+16.2a}{a+a}=\frac{34a}{2a}=17g/mol\)

\(\overline{M_B}=\frac{2b.71+3b.16.2}{2b+3b}=\frac{238b}{5b}=47,6g/mol\)

\(\rightarrow d_{A/B}=\frac{17}{47,6}=\frac{5}{14}\approx0,36\)

Trả lời :

Bạch kim (IV) oxit - PtO₂

Niobi (V) oxit - Nb₂O₅

Photpho pentaflorua  - PF₅

Crom (III) nitrua - CrN

~HT~

CTHH lần lượt là : \(PtO_2,Nb_2O_5,PF_5,CrN\)

Trả lời :

Bạch Kim (IV) oxit - PtO₂

Niobi (V) oxit - Nb₂O₅

Photpho pentaflourua - PF₅

Crom (III) nitrua - CrN

~HT~

H2CO3

h2co3 nha

HCN + NH3 • H2O = NH4CN + H2O

nếu sai thì cho mk xin lỗi nha

\(CaHPO_4+NaHSO_4\rightarrow CaSO_4+NaH_2PO_4\)

Bài 1 :

Gọi 

\(n_{Fe} = a(mol) ; n_{Zn} = b(mol)\\ Fe + 2HCl \to FeCl_2 + H_2\\ Zn + 2HCl \to ZnCl_2 + H_2\\ \)

Ta có : 

\(\hept{\begin{cases}n_{H_2}=a+b=\frac{3,36}{22,4}=0,15\left(mol\right)\\m_{muoi}=127a+136b=19,5\left(gam\right)\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}a=0,1\\b=0,05\end{cases}}\)\(\Rightarrow\hept{\begin{cases}m_{Fe}=0,1.56=5,6\left(gam\right)\\m_{Zn}=0,05.65=3,25\left(gam\right)\end{cases}}\)

Bài 2 : 

\(\hept{\begin{cases}n_{BaCO_3}=a\left(mol\right)\\n_{BaSO_3}=b\left(mol\right)\end{cases}}\)

\(BaCO_3 + 2HCl \to BaCl_2 + CO_2 + H_2O\\ BaSO_3 + 2HCl \to BaCl_2 + SO_2 + H_2O\)

Ta có : 

\(\hept{\begin{cases}m_{hh}=197a+217b=20,5\left(gam\right)\\n_{khí}=n_{CO_2}+n_{SO_2}=a+b=\frac{2,24}{22,4}=0,1\left(mol\right)\end{cases}}\)

Suy ra:  a = 0,06 ; b = 0,04

\(\%m_{BaCO_3} = \dfrac{0,06.197}{20,5}.100\% =57,66\%\\ \%m_{BaSO_3} = 100\%- 57,66\%=42,34\%\)

\(n_{FeCl_2}=0,25mol\)

Đặt \(\hept{\begin{cases}x=n_{Fe}\\y=n_{FeO}\\z=n_{FeCO_3}\end{cases}}\)

Có \(\frac{2x+44z}{x+z}=30\rightarrow-28x+14z=0\)

BT Fe \(x+y+z=0,25\)

Có hệ phương trình

\(\hept{\begin{cases}56x+72y+116z=21,6\\x+y+z=0,25\\-28x+14z=0\end{cases}}\)

\(\rightarrow\hept{\begin{cases}x=0,05\\y=0,1\\z=0,1\end{cases}}\)

\(\%m_{Fe}=\frac{0,05.56}{21,6}.100\%=12,9\%\)

\(\%m_{FeO}=\frac{0,1.72}{21,6}.100\%=33,3\%\)

\(\rightarrow\%m_{FeCO_3}=100\%-12,9\%-33,3\%=53,8\%\)

\(Fe + 2HCl \to FeCl_2 + H_2\\ FeO + 2HCl \to FeCl_2 + H_2O\\ FeCO_3 + 2HCl \to FeCl_2 + CO_2 + H_2O\)

\(n_{FeCl_2}=\frac{31,75}{127}=0,25\left(mol\right)\)

\(\hept{\begin{cases}n_{Fe}=a\\n_{FeO}=b\\n_{FeCO_3}=c\end{cases}\Rightarrow\hept{\begin{cases}56a+72b+116b=21,6\\n_{FeCl_2}=a+b+c=0,25\\\frac{2a+44c}{a+c}=7,5.4\end{cases}}}\)

\(\Rightarrow\hept{\begin{cases}a=0,05\\b=0,1\\c=0,1\end{cases}\Rightarrow\%m_{Fe}=\frac{0,05.56}{21,6}.100\%=12,96\%}\)