=> %m_H2O = 100% - 37,03% = 62,97%

=> m_H2O = \(\dfrac{18a}{23.2+12+16.3+18a}.100\%=62.97\%\)

=> x = 10

=> CTHH: Na₂CO₃ . 10H₂O

\(M_A=17.2=34\left(g/mol\right)\)

\(m_H=\dfrac{5,88.34}{100}=2\left(g\right)\Rightarrow n_H=\dfrac{2}{1}=2\left(mol\right)\)

\(m_S=\dfrac{94,12.34}{100}=32\left(g\right)\Rightarrow n_S=\dfrac{32}{32}=1\left(mol\right)\)

=> CTHH: H₂S

\(M_A=M_{H_2}.17=2.17=34\left(\dfrac{g}{mol}\right)\\ Đặt.CTTQ:H_xS_y\left(x,y:nguyên,dương\right)\\ \Rightarrow x:y=\dfrac{5,88\%}{1}:\dfrac{94,12\%}{32}=5,88\%:2,94\%=2:1\\ \Rightarrow x=2;y=1\\ \Rightarrow CTHH:H_2S\)

\(C+O_2\underrightarrow{t_o}CO_2\)

\(N+O_2\underrightarrow{t_O}NO_2\)

\(4Na+O_2\underrightarrow{t_o}2Na_2O\)

\(2Ca+O_2\underrightarrow{t_O}2CaO\)

\(4Al+3O_2\underrightarrow{t_o}2Al_2O_3\)

\(C_2H_4+3O_2\underrightarrow{t_o}2CO_2+2H_2O\)

\(C_2H_2+\dfrac{5}{2}O_2\underrightarrow{t_o}2CO_2+H_2O\)

1: \(C+O_2\rightarrow CO_2\)

2: \(4N+O_2\rightarrow2N_2O\)

3: \(4Na+O_2\rightarrow2Na_2O\)

4 \(2Ca+O_2\rightarrow2CaO\)

5 \(4Al+3O_2\rightarrow2Al_2O_3\)

6: \(C_2H_4+O_2\rightarrow CH_3CHO\)

7: \(2C_2H_2+5O_2\rightarrow4CO_2+2H_2O\)

1)

PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)

a. \(n_{Zn}=\frac{m}{M}=\frac{13}{65}=0,2mol\)

\(n_{HCl}=\frac{m}{M}=\frac{18,25}{36,5}=0,5mol\)

Lập tỷ lệ \(\frac{n_{Zn}}{1}\) và \(\frac{n_{HCl}}{2}\rightarrow\frac{0,1}{1}< \frac{0,5}{2}\)

Vậy sau phản ứng HCl còn dư nên tính theo số mol Zn

\(\rightarrow n_{Zn}=n_{H_2}=0,2mol\)

\(\rightarrow V_{H_2}=0,2.22,4=4,48l\)

b. Sau phản ứng thì nhúng dung dịch vào quỳ tím, làm cho quỳ tím hoá đỏ bởi còn HCl dư

c. PTHH: \(CuO+H_2\rightarrow^{t^o}Cu+H_2O\)

Trước pứ:    0,3          0,2                                mol

pứ:               0,2          0,2                                mol

Sau pứ:        0,1           0,2                                mol

Vậy sau pứ thu được X gồm CuO dư và Cu

\(\rightarrow m_X=m_{CuO\left(dư\right)}+m_{Cu}=0,1.80+0,2.64=20,8g\)

2)

Đặt \(a\left(g\right)=m_{Na}=m_{Fe}=m_{Al}\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\left(1\right)\)

\(2Na+2HCl\rightarrow2NaCl+H_2\left(2\right)\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\left(3\right)\)

Có \(\hept{\begin{cases}n_{Fe}=a/56mol\\n_{Al}=a/27mol\\n_{Na}=a/23mol\end{cases}}\)

Theo phương trình \(n_{H_2}\left(1\right)=\frac{3}{2}n_{Al}=\frac{a}{18}mol\)

\(n_{H_2}\left(2\right)=\frac{1}{2}n_{Na}=\frac{a}{46}mol\)

\(n_{H_2}\left(3\right)=n_{Fe}=\frac{a}{56}mol\)

\(\frac{a}{18}\approx0,056a\left(mol\right)\)

\(\frac{a}{46}\approx0,22a\left(mol\right)\)

\(\frac{a}{56}\approx0,018a\left(mol\right)\)

Xét \(0,018a< 0,22a< 0,056a\)

Vậy Al cho thể tích \(H_2\) là nhiều nhất.

sao ngọc nam chưa nổi 1000 điểm mà được làm cộng tác viên rồi

\(2HgO\rightarrow\left(450^oC-500^oC\right)2Hg+O_2\uparrow\)

C₂H₄+3O₂-to>2CO₂+2H₂O

              0,3-------0,2-------0,2

n _C2H4=\(\dfrac{4,48}{22,4}\)=0,2 mol

n _O2=\(\dfrac{6,72}{22,4}\)=0,3 mol

=>C2H4 dư

=>V_CO2=V_H2O=0,2.22,4=4,48l

=>V_{C2H4 dư}=0,1.22,4=2,24l

\(n_{O_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ n_{C_2H_4}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ C_2H_4+3O_2\rightarrow\left(t^o\right)2CO_2+2H_2O\\ Vì:\dfrac{0,2}{1}>\dfrac{0,3}{3}\Rightarrow C_2H_4dư\\ \Rightarrow n_{CO_2}=\dfrac{2}{3}.0,3=0,2\left(mol\right)\\ n_{C_2H_4\left(dư\right)}=0,2-\dfrac{0,3}{3}=0,1\left(mol\right)\\ \Rightarrow V_{CO_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\\ V_{C_2H_4\left(dư\right)}=0,1.22,4=2,24\left(l\right)\)

CuO+H₂-to>Cu+H₂O

0,1---------------0,1 mol

n _CuO=\(\dfrac{8}{80}\)=0,1 mol

H=80%

=>m _Cu=0,1.64.\(\dfrac{80}{100}\)=5,12g

Theo PT: \(n_{H_{2\left(lt\right)}}\)

Vì hiệu suất đạt 80%

2KMnO₄-to>K₂MnO₄+MnO₂+O₂

0,2---------------------------------------0,1 mol

n _O2=\(\dfrac{2,24}{22,4}\)=0,1 mol

=>m _{KMnO4 tt} =0,2.158=31,6g

=>H =\(\dfrac{31,6}{39,5}.100\)=80%

Ta có: \(n_{KMnO_4}=\dfrac{39,5}{158}=0,25\left(mol\right)\)

PT: \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)

Theo PT: \(n_{O_2\left(LT\right)}=\dfrac{1}{2}n_{KMnO_4}=0,125\left(mol\right)\)

Mà: \(n_{O_2\left(TT\right)}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

\(\Rightarrow H\%=\dfrac{0,1}{0,125}.100\%=80\%\)

Bạn tham khảo nhé!

\(M=27+\left(14+16.3\right).3=213\left(\dfrac{g}{mol}\right)\)

\(m=n.M=0,15.213=31,95\left(g\right)\)

Đề bài hỏi gì nhỉ?

\(a,Đặt:n_{Mg}=g\left(mol\right);n_{Fe}=j\left(mol\right)\left(g,j>0\right)\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ \Rightarrow\left\{{}\begin{matrix}24g=56j=9,2\\22,4g+22,4j=5,6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}g=0,15\\j=0,1\end{matrix}\right.\\ \Rightarrow\%m_{Mg}=\dfrac{0,15.24}{9,2}.100\approx39,13\%\\ \Rightarrow\%m_{Fe}\approx60,87\%\\ b,n_{HCl}=2n_{H_2}=\dfrac{2.5,6}{22,4}=0,5\left(mol\right)\\ \Rightarrow C_{MddHCl}=\dfrac{0,5}{0,2}=2,5\left(M\right)\\ c,m_{muối}=m_{FeCl_2}+m_{MgCl_2}=95g+127j=95.0,15+127.0,1=26,95\left(g\right)\)

Fe+2HCl->FeCl₂+H₂

x-----------------------x mol

Mg+2HCl->MgCl₂+H₂

y-------------------------y mol

ta có\(\left\{{}\begin{matrix}56x+24y=9,2\\x+y=0,25\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=0,1\\y=0,15\end{matrix}\right.\)

=>%m_Fe=\(\dfrac{0,1.56}{9,2}.100\)=60,87%

=>%m _Mg=39,13%

Ta có : n _HCl=0,1.2+0,15.2=0,5 mol

=>CM_HCl=\(\dfrac{0,5}{0,2}\)=2,5M

=>m _muối =0,1.127+0,15.95=26,95g