khi phân hủy hoàn toàn m(g) kMnO4 thu đc 2,24(l) khí O2( đktc) . tính m=?
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2KMnO4-to>K2MnO4+MnO2+O2
0,3-----------------0,15-----0,15------0,15 mol
n KMnO4=\(\dfrac{47,4}{158}\)=0,3 mol
=>mcr=0,15.197.0,15.87=42,6g
=>VO2=0,15.22,4=3,36l
b) 4P+5O2-to>2P2O5
0,1--------------0,05
nP=\(\dfrac{3,1}{31}\)=0,1 mol
->O2 dư
=>m P2O5=0,05.142=7,1g
mKMnO4 = 47,4/158 = 0,3 (mol)
PTHH: 2KMnO4 -> (t°) K2MnO4 + MnO2 + O2
Mol: 0,3 ---> 0,15 ---> 0,15 ---> 0,15
m = 0,15 . 197 + 0,15 . 87 = 85,2 (g)
V = VO2 = 0,15 . 22,4 = 3,36 (l)
nP = 3,1/31 = 0,1 (mol)
PTHH: 4P + 5O2 -> (t°) 2P2O5
LTL: 0,1/4 < 0,15/5 => O2 dư
nP2O5 = 0,1/2 = 0,05 (mol)
mP2O5 = 0,05 . 142 = 7,1 (g)
a,PTHH: C + O2 -> CO2 (*)
x x x (mol)
S + O2 -> SO2 (**)
y y y (mol)
Ta có dB/H2=9 => MB = 29.2=58 (g)=m hh khí B / n hh Khí B
<=>58= \(\frac{44x+64y}{x+y}\)
=>44x+64y=58x+58y
<=>7x=3y
<=> 7x-3y=0 (1)
Mà hh A =12x+32y=13 (2)
b,Từ (1),(2) ta có hệ pt
\(\hept{\begin{cases}7x-3y=0\\12x+3y=13\end{cases}}\Leftrightarrow\hept{\begin{cases}x=0,15\\y=0,35\end{cases}}\)
=>%C = (12.0,15)/13 .100%=13,84%
%S= 100%-13,84%= 86,16%
Từ pt (*),(**) ta có VO2(đktc)=(x+y).22,4=11,2(l)
2KMnO4-to>K2MnO4+MnO2+O2
1------------------0,5---------0,5----0,5 mol
n O2=\(\dfrac{11,2}{22,4}\)=0,5 mol
=>x =m KMnO4=1.158=158g
=>mA=m K2MnO4+mMnO2=0,5.197+0,5.87=142g
a)
2KMnO4 --to--> K2MnO4 + MnO2 + O2
2Cu + O2 --to--> 2CuO
b)
\(n_{CuO}=\dfrac{16}{80}=0,2\left(mol\right)\)
=> \(n_{O_2}=0,1\left(mol\right)\)
=> \(V_{O_2}=0,1.22,4=2,24\left(l\right)\)
c) \(n_{KMnO_4}=0,2\left(mol\right)\)
=> \(m_{KMnO_4}=0,2.158=31,6\left(g\right)\)
a)
\(2KMnO_4\xrightarrow[]{t^o}K_2MnO_4+MnO_2+O_2\) (1)
\(2KClO_3\xrightarrow[]{t^o}2KCl+3O_2\) (2)
\(n_{KCl}=\dfrac{0,894}{74,5}=0,012\left(mol\right);m_B=\dfrac{0,894}{8,132\%}=11\left(g\right)\)
Gọi \(n_{O_2\left(sinh.ra\right)}=a\left(mol\right)\Rightarrow n_{kk}=3a\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{N_2}=3a.80\%=2,4a\left(mol\right)\\n_{O_2}=a+\left(3a-2,4a\right)=1,6a\left(mol\right)\end{matrix}\right.\)
\(n_C=\dfrac{0,528}{12}=0,044\left(mol\right)\)
\(C+O_2\xrightarrow[]{t^o}CO_2\) (3)
Vì hỗn hợp D gồm 3 khí và O2 chiếm 17,083%
\(\Rightarrow D:CO_2,O_{2\left(d\text{ư}\right)},N_2\)
BTNT C: \(n_{CO_2}=n_C=0,044\left(mol\right)\)
BTNT O: \(n_{O_2\left(d\text{ư}\right)}=n_{O_2\left(b\text{đ}\right)}-n_{CO_2}=1,6a-0,044\left(mol\right)\)
\(\Rightarrow\%V_{O_2}=\%n_{O_2}=\dfrac{1,6a-0,044}{1,6a-0,044+0,044+2,4a}.100\%=17,083\%\)
\(\Leftrightarrow a=0,048\left(mol\right)\left(TM\right)\)
ĐLBTKL: \(m_A=m_B+m_{O_2}=11+0,048.32=12,536\left(g\right)\)
Theo PT (2): \(n_{KClO_3}=n_{KCl}=0,012\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{KClO_3}=\dfrac{0,012.122,5}{12,536}.100\%=11,63\%\\\%m_{KMnO_4}=100\%-11,63\%=88,37\%\end{matrix}\right.\)
b) Theo PT (2): \(n_{O_2}=\dfrac{1}{2}n_{KMnO_4\left(p\text{ư}\right)}+\dfrac{3}{2}n_{KClO_3}\)
\(\Rightarrow n_{KMnO_4\left(p\text{ư}\right)}=2.\left(0,048-\dfrac{3}{2}.0,012\right)=0,06\left(mol\right)\)
\(n_{KMnO_4\left(b\text{đ}\right)}=\dfrac{12,536-0,012.122,5}{158}=0,07\left(mol\right)\)
\(\Rightarrow n_{KMnO_4\left(d\text{ư}\right)}=0,07-0,06=0,01\left(mol\right)\)
\(n_{KCl}=\dfrac{74,5}{74,5}+0,012=1,012\left(mol\right)\)
Theo PT (1): \(n_{K_2MnO_4}=n_{MnO_2}=\dfrac{1}{2}.n_{KMnO_4\left(p\text{ư}\right)}=0,03\left(mol\right)\)
PTHH:
\(2KMnO_4+10KCl+8H_2SO_4\rightarrow6K_2SO_4+2MnSO_4+5Cl_2+8H_2O\) (4)
\(K_2MnO_4+4KCl+4H_2SO_4\rightarrow3K_2SO_4+MnSO_4+2Cl_2+4H_2O\) (5)
\(MnO_2+2KCl+2H_2SO_4\rightarrow MnSO_4+K_2SO_4+Cl_2+2H_2O\) (6)
\(2KCl+H_2SO_4\xrightarrow[]{t^o}K_2SO_4+2HCl\) (7)
Theo PT (4), (5), (6): \(n_{KCl\left(p\text{ư}\right)}=5n_{KMnO_4\left(d\text{ư}\right)}+4n_{K_2MnO_4}+2n_{MnO_2}=0,23\left(mol\right)< 1,012\left(mol\right)=n_{KCl\left(b\text{đ}\right)}\)
`=> KCl` dư
Theo PT (4), (5), (6): \(n_{Cl_2}=\dfrac{1}{2}.n_{KCl\left(p\text{ư}\right)}=0,115\left(mol\right)\)
\(\Rightarrow V_{kh\text{í}}=V_{Cl_2}=0,115.22,4=2,576\left(l\right)\)
Gọi a , b là số mol của KClO3 và KMnO4
TH1: Y có CO2 , N2 , O2 dư
2KClO3 ➝ 2KCl + 3O2
2KMnO4 ➝ K2MnO4 + MnO2 + O2
Gọi nO2 = x => \(\dfrac{nO_{2_{ }}}{_{ }kk}\) = 3x . 0,2 = 0,6x
nN2 = 3x.0,8 = 2,4x
C + O2 ➝ CO2
nCO2 = nC = \(\dfrac{0,528}{12}\) = 0,044
hh khí gồm : nCO2 = 0,044 ; nO2 = 1,6x - 0,044 ; nN2 = 2,4x
=> 0,044 + 1,6x - 0,044 + 2,4x = \(\dfrac{0,044.100}{22,92}\)
<=> x = 0,048
=> mhh đầu = mY + mO2 = \(\dfrac{0,894.100}{8,132}\) + 0,048.32 = 12,53
TH 2 : Y có CO , CO2 ; N2
Bảo toàn C : nCO + nCO2 = nC = 0,044 => nCO = 0,044 - nCO2
Bảo toàn O : 0,5.nCO + nCO2 = nO2 = 1,6a
⇒ 0,5.( 0,044 - nCO2 ) + nCO2 = 1,6a => nCO2 = 3,2a - 0,044
Tổng mol hh : nCO + nCO2 + nN2 = 0,044 + 2,4a
=> \(\dfrac{3,2a-0,044}{0,044+2,4a}\) = \(\dfrac{22,92}{100}\)
a ≈0.02
=> m = m rắn + mO2 = \(\dfrac{0,894.100}{8,132}\) + 0,02 . 32 = 11,646 ( g )
a)nO2=\(\dfrac{3.36}{22.4}\)=0,15(mol)
2KMnO4(to)→K2MnO4+MnO2+O2
Theo PT: nKMnO4=2nO2=0,3(mol)
→m=mKMnO4=0,3.158=47,4(g)
b)nH2=\(\dfrac{8.96}{22.4}\)=0,4(mol)
2H2+O2(to)→2H2O
Vì \(\dfrac{nH_2}{2}\)<nO2→O2nH2 dư
Theo PT: nH2O=nH2=0,4(mol)
→mH2O=0,4.18=7,2(g)
2KMnO4-to>K2MnO4+MnO2+O2
0,2--------------------------------------0,1 mol
n O2=\(\dfrac{2,24}{22,4}\)=0,1 mol
=>m KMnO4=0,2.158=31,6g
nO2 = 2,24/22,4 = 0,1 (mol)
PTHH: 2KMnO4 -> (t°) K2MnO4 + MnO2 + O2
Mol: 0,2 <--- 0,1 <--- 0,1 <--- 0,1
mKMnO4 = 0,2 . 158 = 31,6 (g)