Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
nC = 1,8 : 12 = 0,15 (mol)
pthh : C+O2 --> CO2
0,15>0,15 (mol)
=> V O2 = 0,15 .22,4 = 3,36 (l)
=> Vkk = 3,36 : 1/5 = 16,8 (L)
nZn = 13 : 65 = 0,2 (mol)
pthh : 2Zn + O2 -t-> 2ZnO
0,2-----> 0,1 (mol)
=>VO2 = 0,1.22,4 = 2,24 (l)
=> Vkk = 2,24 : 1/5 = 11,2 (l)
nAl = 2,7 : 27 = 0,1 (mol)
pthh : 4Al + 3O2 --t--->2 Al2O3
0,1-->0,075 (mol)
=> VO2 = 0,075 . 22,4 = 1, 68 (l)
=> VKk = 1,68 : 1/5 = 8,4 (l)
a, nC = 1,8/12 = 0,15 (mol)
PTHH: C + O2 -> (t°) CO2
Mol: 0,15 ---> 0,3
Vkk = 0,3 . 5 . 22,4 = 33,6 (l)
b, nZn = 13/65 = 0,2 (mol)
PTHH: 2Zn + O2 -> (t°) 2ZnO
Mol: 0,2 ---> 0,1
Vkk = 0,1 . 5 . 22,4 = 11,2 (l)
c, nAl = 2,7/27 = 0,1 (mol)
PTHH: 2Al + 3O2 -> (t°) 2Al2O3
Mol: 0,1 ---> 0,075
Vkk = 0,075 . 5 . 22,4 = 8,4 (l)
\(1,2H_2+O_2\underrightarrow{t}2H_2O\)
\(2Mg+O_2\underrightarrow{t}2MgO\)
\(2Cu+O_2\underrightarrow{t}2CuO\)
\(S+O_2\underrightarrow{t}SO_2\)
\(4Al+3O_2\underrightarrow{t}2Al_2O_3\)
\(C+O_2\underrightarrow{t}CO_2\)
\(4P+5O_2\underrightarrow{t}2P_2O_5\)
\(2,PTHH:C+O_2\underrightarrow{t}CO_2\)
\(a,n_{O_2}=0,2\left(mol\right)\Rightarrow n_{CO_2}=0,2\left(mol\right)\Rightarrow m_{CO_2}=8,8\left(g\right)\)
\(b,n_C=0,3\left(mol\right)\Rightarrow n_{CO_2}=0,3\left(mol\right)\Rightarrow m_{CO_2}=13,2\left(g\right)\)
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.
\(a,n_{O_2}\left(dư\right)=0,05\left(mol\right)\Rightarrow m_{O_2}\left(dư\right)=1,6\left(g\right)\)
\(b,n_{P_2O_5}=0,1\left(mol\right)\Rightarrow m_{P_2O_5}=14,2\left(g\right)\)
4P+5O2-to>2P2O5
0,1----0,125 mol
n P=\(\dfrac{3,1}{31}\)=0,1 mol
=>Vkk=0,125.22,4.5=14l
=>D
\(n_{P_2O_5}=\dfrac{21,3}{142}=0,15\left(mol\right)\)
PTHH: 4P + 5O2 --to--> 2P2O5
0,3 0,375 0,15
\(\rightarrow\left\{{}\begin{matrix}m_P=0,3.31=9,3\left(g\right)\\V_{O_2}=0,375.22,4=8,4\left(l\right)\\V_{kk}=8,4.5=42\left(l\right)\end{matrix}\right.\)
PTHH: 2KClO3 --to--> 2KCl + 3O2
0,25 0,375
=> mKClO3 = 0,25.122,5 = 30,625 (g)
\(nP_2O_5=\dfrac{21,3}{142}=0,15\left(mol\right)\)
\(pthh:4P+5O_2-t^o->2P_2O_5\)
0,3 0,375 0,15
=> \(m_P=0,3.31=9,3\left(g\right)\)
=>\(V_{O_2}=0,375.22,4=8,4\left(L\right)=>V_{KK}=8,4:20\%=42\left(L\right)\)
\(pthh:2KMnO_4-t^o->K_2MnO_4+MnO_2+O_2\)
0,75 0,75
=> mKMnO4 = 0,75 . 158 = 118,5 (G)
a)
nP =62 : 31 = 2 (mol)
PTHH:4P + 5O2 --(to)-> 2P2O5
Theo PTHH: \(nO_2=\dfrac{5}{4}nP=\dfrac{5}{4}.2=2,5\left(mol\right)\)
VO2(đktc) = 2,5 ×22,4=56 (lít)
\(\dfrac{100\%}{21\%}.56=227\left(lít\right)\)
b)
\(nP=\dfrac{15,5}{31}=0,5\left(mol\right)\)
\(nO_2=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)
\(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
0,4 0,5 0,2
tính theo pthh : => P dư , O2 đủ
nP(dư) = 0,5-0,4=0,1(mol)
=> mP (dư) = 0,1 . 31 = 3,1(g)
mP2O5 = 0,5 . 142=71(g)
a) PTHH: 4P+5O2-----to---> 2P2O5
0,2 0,25 0,1
b)\(n_{P_2O_5}=\dfrac{m}{M}=\dfrac{14,2}{142}=0,1\left(mol\right)\)
\(m_P=n.M=0,2.31=6,2\left(gam\right)\)
c) \(V_{O_2}=n.22,4=0,25.22,4=5,6\left(l\right)\)
a, \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
PT: \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
Theo PT: \(n_{O_2}=\dfrac{3}{4}n_{Al}=0,15\left(mol\right)\Rightarrow V_{O_2}=0,15.24,79=3,7185\left(l\right)\)
\(\Rightarrow V_{kk}=\dfrac{V_{O_2}}{20\%}=18,5925\left(l\right)\)
b, Theo PT: \(n_{Al_2O_3}=\dfrac{1}{2}n_{Al}=0,1\left(mol\right)\Rightarrow m_{Al_2O_3}=0,1.102=10,2\left(g\right)\)
\(n_P=\dfrac{3,1}{31}=0,1\left(mol\right)\)
PTHH: 4P + 5O2 ---to→ 2P2O5
Mol: 0,1 0,125
\(V_{O_2}=0,125.22,4=2,8\left(l\right)\Rightarrow V_{kk}=\dfrac{2,8}{20}.100=14\left(l\right)\)
a, \(PTHH:4P+5O_2\rightarrow2P_2O_5\)
\(n_P=\frac{46,5}{31}=1,5\left(mol\right)\)
\(\Rightarrow n_{O2}=1,875\left(mol\right)\)
\(\Rightarrow V_{O2}=1,875.22,4=42\left(l\right)\)
\(\Rightarrow V_{kk}=42.5=210\left(l\right)\)
b,\(PTHH:C+O_2\rightarrow CO_2\)
\(n_C=\frac{30}{12}=2,5\left(mol\right)\)
\(\Rightarrow n_{O2}=n_C=2,5\left(mol\right)\)
\(\Rightarrow V_{O2}=2,5.22,4=56\left(l\right)\)
\(\Rightarrow V_{kk}=56.5=280\left(l\right)\)
c,\(PTHH:4Al+3O_2\rightarrow2Al_2O_3\)
\(n_{Al}=\frac{67,5}{27}=2,5\left(mol\right)\)
\(\Rightarrow n_{O2}=1,875\left(mol\right)\)
\(\Rightarrow V_{O2}=1,875.22,4=42\left(l\right)\)
\(\Rightarrow V_{kk}=42.5=210\left(l\right)\)
d,\(PTHH:2H_2+O_2\rightarrow2H_2O\)
\(n_{H2}=\frac{33,6}{22,4}=1,5\left(mol\right)\)
\(\Rightarrow n_{O2}=0,75\left(mol\right)\)
\(\Rightarrow V_{O2}=0,75.22,4=16,8\left(l\right)\)
\(\Rightarrow V_{kk}=16,8.5=84\left(l\right)\)