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a) Phương trình hóa học chữ:
Kali permanganat =(nhiệt)=> Kali manganat + Mangan dioxit + Oxi
b) Áp dụng định luật bảo toàn khối lượng
=> mKMnO4 = mK2MnO4 + mMnO2 + mO2 = 197 + 87 + 32 = 316 gam
=> Phần trăm theo khối lượng của KMnO4 trong thuốc tím là:
%mKMnO4 = \(\frac{316}{350}.100\%=90,3\%\)
\(1,2H_2+O_2\underrightarrow{t}2H_2O\)
\(2Mg+O_2\underrightarrow{t}2MgO\)
\(2Cu+O_2\underrightarrow{t}2CuO\)
\(S+O_2\underrightarrow{t}SO_2\)
\(4Al+3O_2\underrightarrow{t}2Al_2O_3\)
\(C+O_2\underrightarrow{t}CO_2\)
\(4P+5O_2\underrightarrow{t}2P_2O_5\)
\(2,PTHH:C+O_2\underrightarrow{t}CO_2\)
\(a,n_{O_2}=0,2\left(mol\right)\Rightarrow n_{CO_2}=0,2\left(mol\right)\Rightarrow m_{CO_2}=8,8\left(g\right)\)
\(b,n_C=0,3\left(mol\right)\Rightarrow n_{CO_2}=0,3\left(mol\right)\Rightarrow m_{CO_2}=13,2\left(g\right)\)
c, Vì\(\frac{0,3}{1}>\frac{0,2}{1}\)nên C phản ửng dư, O2 phản ứng hết, Bài toán tính theo O2
\(n_{O_2}=0,2\left(mol\right)\Rightarrow n_{CO_2}=0,2\left(mol\right)\Rightarrow m_{CO_2}=8,8\left(g\right)\)
\(3,PTHH:CH_4+2O_2\underrightarrow{t}CO_2+2H_2O\)
\(C_2H_2+\frac{5}{2}O_2\underrightarrow{t}2CO_2+H_2O\)
\(C_2H_6O+3O_2\underrightarrow{t}2CO_2+3H_2O\)
\(4,a,PTHH:4P+5O_2\underrightarrow{t}2P_2O_5\)
\(n_P=1,5\left(mol\right)\Rightarrow n_{O_2}=1,2\left(mol\right)\Rightarrow m_{O_2}=38,4\left(g\right)\)
\(b,PTHH:C+O_2\underrightarrow{t}CO_2\)
\(n_C=2,5\left(mol\right)\Rightarrow n_{O_2}=2,5\left(mol\right)\Rightarrow m_{O_2}=80\left(g\right)\)
\(c,PTHH:4Al+3O_2\underrightarrow{t}2Al_2O_3\)
\(n_{Al}=2,5\left(mol\right)\Rightarrow n_{O_2}=1,875\left(mol\right)\Rightarrow m_{O_2}=60\left(g\right)\)
\(d,PTHH:2H_2+O_2\underrightarrow{t}2H_2O\)
\(TH_1:\left(đktc\right)n_{H_2}=1,5\left(mol\right)\Rightarrow n_{O_2}=0,75\left(mol\right)\Rightarrow m_{O_2}=24\left(g\right)\)
\(TH_2:\left(đkt\right)n_{H_2}=1,4\left(mol\right)\Rightarrow n_{O_2}=0,7\left(mol\right)\Rightarrow m_{O_2}=22,4\left(g\right)\)
\(5,PTHH:S+O_2\underrightarrow{t}SO_2\)
\(n_{O_2}=0,46875\left(mol\right)\)
\(n_{SO_2}=0,3\left(mol\right)\)
Vì\(0,46875>0,3\left(n_{O_2}>n_{SO_2}\right)\)nên S phản ứng hết, bài toán tính theo S.
\(a,\Rightarrow n_S=n_{SO_2}=0,3\left(mol\right)\Rightarrow m_S=9,6\left(g\right)\)
\(n_{O_2}\left(dư\right)=0,16875\left(mol\right)\Rightarrow m_{O_2}\left(dư\right)=5,4\left(g\right)\)
\(6,a,PTHH:C+O_2\underrightarrow{t}CO_2\)
\(n_{O_2}=1,5\left(mol\right)\Rightarrow n_C=1,5\left(mol\right)\Rightarrow m_C=18\left(g\right)\)
\(b,PTHH:2H_2+O_2\underrightarrow{t}2H_2O\)
\(n_{O_2}=1,5\left(mol\right)\Rightarrow n_{H_2}=0,75\left(mol\right)\Rightarrow m_{H_2}=1,5\left(g\right)\)
\(c,PTHH:S+O_2\underrightarrow{t}SO_2\)
\(n_{O_2}=1,5\left(mol\right)\Rightarrow n_S=1,5\left(mol\right)\Rightarrow m_S=48\left(g\right)\)
\(d,PTHH:4P+5O_2\underrightarrow{t}2P_2O_5\)
\(n_{O_2}=1,5\left(mol\right)\Rightarrow n_P=1,2\left(mol\right)\Rightarrow m_P=37,2\left(g\right)\)
\(7,n_{O_2}=5\left(mol\right)\Rightarrow V_{O_2}=112\left(l\right)\left(đktc\right)\);\(V_{O_2}=120\left(l\right)\left(đkt\right)\)
\(8,PTHH:C+O_2\underrightarrow{t}CO_2\)
\(m_C=0,96\left(kg\right)\Rightarrow n_C=0,08\left(kmol\right)=80\left(mol\right)\Rightarrow n_{O_2}=80\left(mol\right)\Rightarrow V_{O_2}=1792\left(l\right)\)
\(9,n_p=0,2\left(mol\right);n_{O_2}=0,3\left(mol\right)\)
\(PTHH:4P+5O_2\underrightarrow{t}2P_2O_5\)
Vì\(\frac{0,2}{4}< \frac{0,3}{5}\)nên P hết O2 dư, bài toán tính theo P.
\(a,n_{O_2}\left(dư\right)=0,05\left(mol\right)\Rightarrow m_{O_2}\left(dư\right)=1,6\left(g\right)\)
\(b,n_{P_2O_5}=0,1\left(mol\right)\Rightarrow m_{P_2O_5}=14,2\left(g\right)\)
2KClO3\(\rightarrow\) 2KCl + 3O2
\(\text{nO2(đktc) = 53,76 : 22,4 = 2,4 (mol)}\)
\(\rightarrow\) mO2 = nO2.MO2 = 2,4.32 = 76,8 (g)
BTKL ta có: mKClO3 bđ = m rắn + mO2 = 168,2 + 76,8=245 (g)
b) 2KMnO4\(\rightarrow\) K2MnO4 + MnO2 + O2
4,8 ________________________2,4 (mol)
Theo PTHH: \(\text{nKMnO4 = 2nO2 = 2.2,4 = 4,8 (mol)}\)
\(\rightarrow\) mKMnO4 lí thuyết = 4,8.158 = 758,4 (g)
Vì %H = 90% nên
\(\text{mKMnO4 thực tế cần lấy = mKMnO4 lí thuyết.100%:90% = 758,4.100%:90%= 482,67 (g)}\)
a, \(3Fe+2O_2\underrightarrow{^{to}}Fe_3O_4\)
\(n_{Fe3O4}=\frac{4,64}{232}=0,02\left(mol\right)\)
\(\rightarrow n_{Fe}=3n_{Fe3O4}=0,06\left(mol\right)\)
\(n_{O2}=2n_{Fe3O4}=0,04\left(mol\right)\)
b,\(PTHH:2KMnO_4\underrightarrow{^{to}}KMnO_2+MnO_2+O_2\)
\(n_{KMnO4}=2n_{O2}=0,08\left(mol\right)\)
\(\rightarrow m_{KMnO4O}=0,08.158=12,64\left(g\right)\)
1.
\(2KClO_3\underrightarrow{^{to}}2KCl+3O_2\)
\(n_{KClO3}=\frac{9,8}{122,5}=0,08\left(mol\right)\)
\(\Rightarrow n_{O2}=\frac{3}{2}n_{KClO3}=\frac{3}{2}.0,08=0,12\left(mol\right)\)
\(\Rightarrow V_{O2}=0,12.22,4=2,688\left(l\right)\)
2.
\(a,2KMnO_4\underrightarrow{^{to}}K_2MnO_4+MnO_2+O_2\)
\(b,n_{O2}=\frac{33,6}{22,4}=1,5\left(mol\right)\)
\(\Rightarrow n_{KMnO4}=2n_{O2}=2.1,5=3\left(mol\right)\)
\(\Rightarrow m_{KMnO4}=3.158=474\left(g\right)\)
3.
\(2KMnO_4\underrightarrow{^{to}}K_2MnO_4+MnO_2+O_2\)
1____________________________0,5
\(2KClO_3\underrightarrow{^{to}}2KCl+3O_2\left(1\right)\)
1____________________1,5
Đặt \(n_{KMnO4}=n_{KClO3}=1\left(mol\right)\)
\(V_{O2\left(1\right)}=0,5.22,4=11,2\left(l\right)\)
\(V_{O2\left(2\right)}=1,5.22,4=33,6\left(l\right)\)
Vậy nung KClO3 sẽ cho thể tích oxi nhiều hơn.
2.
PTHH: CuO + H2 \(\rightarrow\) Cu + H2O
ADCT: n = \(\dfrac{m}{M}\) ta có:
nCuO = 24/80 = 0,3 (mol)
Theo PTHH : nH2 = nCuO = 0,3 (mol)
ADCT: V= 22,4 . n
VH2 = 22,4 . 0,3 = 6,72 (l)
3. PTHH: 3Fe + 2O2 \(\rightarrow\) Fe3O4
ADCT: n = \(\dfrac{m}{M}\)ta có:
nFe = 16,8/56 = 0,3 (mol)
Theo PTHH: nO2 = nFe = 0,3 (mol)
ADCT: V = 22,4 . n
VO2= 22,4. 0,3 = 6,72 (l)
a,4P+5O2→t02P2O5
b,nP=mM=6,232=0,19375(mol)
Theo PTHH :
nO2=54nP=54.0,19375=0,24(mol)
⇒VO2=n.22,4=0,24.22,4=5,376(l)
c, Theo PTHH :
nP2O5=12nP=12.0,19375=0,097(mol)
⇒mP2O5=n.M=0,097.142=13,774(g)
d, 2KMnO4→K2MnO4+MnO2+O2
Theo PTHH :
nKMnO4=2nO2=2.0,24=0,48(mol)
⇒mKMnO4=n.M=0,48.158=75,84(g)
a,\(4P+5O_2\rightarrow^{t^0}2P_2O_5\)
\(b,n_P=\dfrac{m}{M}=\dfrac{6,2}{32}=0,19375\left(mol\right)\)
Theo PTHH :
\(n_{O_2}=\dfrac{5}{4}n_P=\dfrac{5}{4}.0,19375=0,24\left(mol\right)\)
\(\Rightarrow V_{O_2}=n.22,4=0,24.22,4=5,376\left(l\right)\)
c, Theo PTHH :
\(n_{P_2O_5}=\dfrac{1}{2}n_P=\dfrac{1}{2}.0,19375=0,097\left(mol\right)\)
\(\Rightarrow m_{P_2O_5}=n.M=0,097.142=13,774\left(g\right)\)
d, \(2KMnO_4\rightarrow K_2MnO_4+MnO_2+O_2\)
Theo PTHH :
\(n_{KMnO_4}=2n_{O_2}=2.0,24=0,48\left(mol\right)\)
\(\Rightarrow m_{KMnO_4}=n.M=0,48.158=75,84\left(g\right)\)
\(n_{O_2}=\dfrac{6,72}{5.22,4}=0,06\left(mol\right)\\ PTHH:2Zn+O_2\underrightarrow{t^o}2ZnO\\ Mol:0,03\leftarrow0,06\rightarrow0,03\\ \rightarrow\left\{{}\begin{matrix}a=0,03.65=1,95\left(g\right)\\x=0,03.81=2,43\left(g\right)\end{matrix}\right.\)
\(PTHH:2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\uparrow\)
0,12 0,06
\(\rightarrow m_{KMnO_4}=0,12.158=18,96\left(g\right)\)
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