Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) PTHH :
2KClO3 \(\rightarrow\)2KCl + 3O2 (1)
CH4 + 2O2 \(\rightarrow\)CO2 + 2H2O (2)
2H2 + O2 \(\rightarrow\) 2H2O (3)
Theo PT(2) => nO2 = 2.nCH4 = 2 x 0.5 =1(mol)
Theo PT(3) => nO2 = 1/2 x nH2 = 1/2 x 0.25 =0.125(mol)
=> tổng nO2 = 1+ 0.125 =1.125(mol)
Theo PT(1) => nKClO3 = 2/3 . nO2 = 2/3 x 1.125 = 0.75(mol)
=> mKClO3 = n .M = 0.75 x 122.5 =91.875(g)
b) 4Al + 3O2 \(\rightarrow\) 2Al2O3 (4)
2Zn + O2 \(\rightarrow\) 2ZnO (5)
nAl = m : M = 6.75/27=0.25(mol)
nZn = m/M = 9.75/65 =0.15(mol)
Theo PT(4) => nO2 = 3/4 . nAl = 3/4 x 0.25 =0.1875(mol)
Theo PT(5) => nO2 = 1/2 x nZn = 1/2 x 0.15 =0.075(mol)
tổng nO2 = 0.1875 + 0.075 =0.2625(mol)
theo PT(1) => nKClO3 = 2/3 x nO2 = 2/3 x 0.2625 =0.175(mol)
=> mKClO3 = n .M = 0.175 x 122.5 =21.4375(g)
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)
Áp dụng ĐLBTKL ta có:
\(m_{hh}+m_{O_2}=m_{hhoxit}\)
\(\Rightarrow m_{O_2}=m_{hhoxit}-m_{hh}=26,21-17,49=8,72\left(g\right)\)
\(n_{O_2}=\frac{8,72}{32}=0,2725\left(mol\right)\rightarrow V_{O_2}=22,4.0,2725=6,104\left(l\right)\)
Đặt \(n_{Al}=x\left(mol\right);n_{Fe}=y\left(mol\right)\)
\(PTHH:4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
(mol)_____x___________0,5x
\(PTHH:3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
(mol)_____y__________y/3
Theo đề bài ta có:
\(\left\{{}\begin{matrix}27x+56y=17,49\\102.0,5x+\frac{232.y}{3}=26,21\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,15\\y=0,24\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\frac{0,15.27}{17,49}.100\%=23,2\left(\%\right)\\\%m_{Fe}=100-23,2=76,8\left(\%\right)\end{matrix}\right.\)
a) 3Fe+2O2--->FE3O4
0,5------1/3 (mol)
4Al+3O2---.2Al2O3
1,25--0,9375(mol)
2Zn+O2--->2ZnO
1,5---0,75(mol)
n O2=1/3+0,9375+0,75=2,02(mol)
m O2=2,02.32=64,64(g)
b) 4P+5O2-->2P2O5
0,1-----0,125(mol)
S+02--->SO2
0,2--0,2(mol)
C+O2-->CO2
0,3--0,3(mol)
n O2=0,125+0,2+0,3=0,625(mol)
m O2=0,625.32=20(g)
c) n CH4=1,6/16=0,1(mol)
n CO=2,8/28=0,1(mol)
n C4H10=0,58/58=0,01(mol)
CH4+2O2--->CO2+2H2O
0,1---0,2(mol)
2CO+O2-->2CO2
0,1--0,05(mol)
C4H10+13/2O2--->4CO2+5H2O
0,01-----0,065(mol)
n O2=0,2+0,05+0,065=0,315(mol)
m O2=0,315.32=10,08(g)
Bạn lập các PTHH sau đó chuyển tất cả thành số mol.Tính số mol của Oxi theo các chất đã cho
a, \(PTHH:2KClO_3\underrightarrow{^{to}}2KCl+3O_2\)
__________0,5____________0,75 (mol)
\(\rightarrow m_{O2}=0,075.32=24\left(g\right)\)
b, \(PTHH:2KNO_3\underrightarrow{^{to}}2KNO_2+O_2\uparrow\)
__________0,5________________0,25 (mol)
\(\rightarrow m_{O2}=0,25.32=8\left(g\right)\)
c, \(PTHH:2KClO_3\underrightarrow{^{to}}2KCl+3O_2\uparrow\)
\(n_{KClO3}=\frac{2,45}{122,5}=0,02\left(mol\right)\)
\(2KClO_3\underrightarrow{^{to}}2KCl+3O_2\uparrow\)
0,2_____________0,03 (mol)
\(\rightarrow m_{O2}=0,03.32=0,96\left(g\right)\)
d, \(n_{KnO3}=\frac{24,5}{101}=0,24\left(mol\right)\)
\(PTHH:2KNO_3\underrightarrow{^{to}}2KNO_2+O_2\)
__________0,24______________0,12 (mol)
\(\rightarrow m_{O2}=0,12.32=3,84\left(g\right)\)
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)
=> mKClO3= 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