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)

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

Đáp án D

n_H2O = 4,5 : 18 = 0,25 (mol)

n_HCHO = ( 0,15.2 + 0,25 – 0, 25.2) = 0,05 (mol) (Bảo toàn nguyên tố O)

n_C2H2 = ( 0,15 – 0,05 )/ 2 = 0,05 (mol) (Bảo toàn nguyên tố C)

n_H2 = ( 0,25.2 – 0,05.2 – 0,05.2) / 2 = 0,15 (mol) (Bảo toàn nguyên tố H)

%C₂H₂ = [(0,05.26)/ ( 0,05.30 + 0,05.26 + 0,15.2)].100% = 41,935%

Đáp án D

=> x = 0,05 mol

Đáp án D