a) \(CH_4+2O_2\underrightarrow{t\text{°}}CO_2+2H_2O\)(1)
___0,5------>1___________________(mol)

\(2H_2+O_2\underrightarrow{t\text{°}}2H_2O\)(2)
0,25->0,125__0,25____(mol)

\(2KClO_3\underrightarrow{t\text{°}}2KCl+3O_2\)(3)
0,75<---------------0,125+1=1,125(mol)

=> m_KClO3= 0,75*122,5=91,875(g)

b) n _Al= 6,75/27=0,25(mol)

n _Zn= 9,75/65= 0,15 (mol)

\(4Al+2O_2\underrightarrow{t\text{°}}2Al_2O_3\) (1')
0.25-->0.125_________(mol)

\(2Zn+O_2\underrightarrow{t\text{​​}\text{°}}2ZnO\) (2')
0.15->0.075_________(mol)

\(2KClO_3\underrightarrow{t\text{°}}2KCl+3O_2\)(3')
\(\frac{2}{15}\)<-------------------0.075+0.125=0.2(mol)

m _KClO3=\(\frac{2}{15}\)*122,5=\(\frac{49}{3}\) (g)

tại sao câu b chỗ PT Al í lại là 2O2 phải là 3O2 chứ

b, \(4P+5O_2\underrightarrow{t^o}2P_2O_5\)

\(n_P=\dfrac{6,2}{31}=0,2mol\\n_{O_2}=\dfrac{0,2.5}{4}=0,25mol \)

\(S+O_2\underrightarrow{t^o}SO_2\)

\(n_S=\dfrac{3,2}{32}=0,1mol\\ n_{O_2}=0,1mol\)

\(C+O_2\underrightarrow{t^o}CO_2\)

\(n_C=\dfrac{2,4}{12}=0,2mol\\ n_{O_2}=0,2mol\\ n_{O_2}\left(tổng\right)=\)

\(0,25+0,1+0,2=0,55mol\\ m_{O_2}\left(trong.hh.B\right)=0,55.32=17,6g\)

a, \(m_{Fe}=0,25.56=14g\)

\(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)

\(\Rightarrow n_{O_2}=\dfrac{0,25.2}{3}=0,16mol\\ m_{O_2}=0,16.32=5,12g\)

\(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)

\(n_{O_2}=\dfrac{0,25.3}{4}=0,1875mol\\ m_{O_2}=0,1875.32=6g\)

\(2Zn+O_2\underrightarrow{t^o}2ZnO\)

\(n_{O_2}=\dfrac{0,5.1}{2}=0,25mol\\ m_{O_2}=0,25.32=8g\)

\(\Rightarrow m_{O_2}\left(trong.hỗn.hợp.A\right)=\) \(5,12+6+8=19,12g\)

nAl= 6,75/27=0,25(mol)

nZn= 9,75/65= 0,15(mol)

4 Al + 3 O2 -to-> 2 Al2O3

0,25____0,1875___0,125(mol)

Zn + 1/2 O2 -to->ZnO

0,15___0,075___0,15(mol)

=> n(O2, tổng)= 0,1875+ 0,075= 0,2625(mol)

PTHH: 2 KMnO4 -to-> K2MnO4 + MnO2 + O2

0,525<----------------------------------------------0,2625(mol)

=> mKMnO4= 0,525.158= 82,95(g)

=> m=82,95(g)

Giải chi tiết cho mik nhé!

\(n_C=\dfrac{1.2}{12}=0.1\left(mol\right)\\ n_S=\dfrac{4}{32}=0.125\left(mol\right)\)

\(2KMnO_4\underrightarrow{t^0}K_2MnO_4+MnO_2+O_2\)

\(C+O_2\underrightarrow{t^0}CO_2\)

\(S+O_2\underrightarrow{t^0}SO_2\)

\(\sum n_{O_2}=n_C+n_S=0.1+0.125=0.225\left(mol\right)\)

\(\Rightarrow n_{KMnO_4}=2n_{O_2}=0.225\cdot2=0.45\left(mol\right)\)

\(m_{KMnO_4}=0.45\cdot158=71.1\left(g\right)\)

2KMnO4 -to>K2MnO4+O2+MnO2

0,45------------------------0,225

C+O2-to>CO2

0,1--0,1

S+O2-to>SO2

0,125-0,125

n C=1,2\12=0,1 mol

n S=4\32=0,125 mol

=>m KMnO4=0,45.158=77,1g

Bài 1 :

\(n_{Na}=\dfrac{m}{M}=0,1\left(mol\right)\)

\(4Na+O_2\rightarrow2Na_2O\)

..0,1....0,025....0,05.......

a, \(V_{O_2}=n.22,4=0,56\left(l\right)\)

b, \(m=m_{Na_2o}=n.M=3,1\left(g\right)\)

Bài 2 :

\(n_{Al}=\dfrac{m}{M}=0,1\left(mol\right)\)

\(4Al+3O_2\rightarrow2Al_2O_3\)

..0,1...0,075...

\(\Rightarrow n_{O_2}=0,075\left(mol\right)\)

Mà : \(\Sigma n_{O_2}=\dfrac{V}{22,4}=0,4\left(mol\right)\)

\(\Rightarrow n_{O_2\left(Mg\right)}=0,4-0,075=0,325\left(mol\right)\)

\(2Mg+O_2\rightarrow2MgO\)

.0,65.....0,325........

\(\Rightarrow m_{Mg}=15,6\left(g\right)\)

\(\Rightarrow m_{hh}=2,7+15,6=18,3\left(g\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%Al=~14,75\\\%Mg=~85,25\end{matrix}\right.\) %

Bài 3 :

- Gọi số mol Al và Mg lần lượt là x , y

\(4Al+3O_2\rightarrow2Al_2O_3\)

..x....0,75x

\(2Mg+O_2\rightarrow2MgO\)

..y........0,5y...........

Có : \(n_{O_2}=0,75x+0,5y=\dfrac{V}{22,4}=0,1\left(mol\right)\left(I\right)\)

Lại có : \(m_{hh}=m_{Al}+m_{Mg}=27x+24y=3,9\left(II\right)\)

- Giair ( i ) và ( ii ) ta được : \(\left\{{}\begin{matrix}x=0,1\\y=0,05\end{matrix}\right.\) ( mol )

\(\Rightarrow\left\{{}\begin{matrix}\%Al=~69,23\\\%Mg=~30,77\end{matrix}\right.\) %

Vậy ...

2Zn+O2-to>2ZnO

0,05--0,025---0,05

n Zn=0,05 mol

=>=>VO2=0,025.22,4=0,56l

2KClO3-to->2KCl+3O2

1\60------------------------0,025

=>m KClO3=\(\dfrac{1}{60}\).122,5=2,041g

\(n_{Zn}=\dfrac{3,25}{65}=0,05mol\)

\(2Zn+O_2\underrightarrow{t^o}2ZnO\)

0,05    0,025

\(V_{O_2}=0,025\cdot22,4=0,56l\)

\(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)

\(\dfrac{1}{60}\)                           0,025

\(m_{KClO_3}=\dfrac{1}{60}\cdot122,5=2,042g\)

\(n_{Al}=\dfrac{12,15}{27}=0,45\left(mol\right)\\ 4Al+3O_2\rightarrow\left(t^o\right)2Al_2O_3\\ n_{O_2\left(LT\right)}=\dfrac{3}{4}.0,45=\dfrac{27}{80}\left(mol\right)\\ n_{O_2\left(ban.đầu\right)}=\dfrac{27}{80}.\left(100\%+10\%\right)=\dfrac{297}{800}\left(mol\right)\\ Gọi:n_{KMnO_4}=a\left(mol\right);n_{KClO_3}=b\left(mol\right)\left(a,b>0\right)\\ 2KMnO_4\rightarrow\left(t^o\right)K_2MnO_4+MnO_2+O_2\\ 2KClO_3\rightarrow\left(t^o\right)2KCl+3O_2\\ \Rightarrow\left\{{}\begin{matrix}158a+122,5b=56,1\\0,8.0,5a+0,85.1,5b=\dfrac{297}{800}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,17087\\b=0,23757\end{matrix}\right.\\ \)

\(n_{Al_2O_3}=\dfrac{0,45}{2}=0,225\left(mol\right)\\ \Rightarrow m_{Al_2O_3}=102.0,225=22,95\left(g\right)\\ m_{chất.còn.lại}=m_{Al_2O_3}+m_{KMnO_4\left(còn\right)}+m_{KClO_3\left(còn\right)}\\ \approx22,95+0,2.0,17087.158+0,15.0,23757.122,5\approx32,715\left(g\right)\)