Bài 7: Tính thể tích không khí cần đốt cháy hoàn toàn:
a) 2,479 lít khí H2 (đkc).
b) 12 gam sulfur.
c) 69 gam sodium.
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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.\(n_{Fe_3O_4}=\dfrac{2,32}{232}=0,01mol\)
\(3Fe+2O_2\rightarrow\left(t^o\right)Fe_3O_4\)
0,3 0,1 ( mol )
\(m_{Fe}=0,3.56=16,8g\)
2.\(n_{Cu}=\dfrac{3,2}{64}=0,05mol\)
\(2Cu+O_2\rightarrow\left(t^o\right)2CuO\)
0,05 0,05 ( mol )
\(m_{CuO}=0,05.80=4g\)
3.\(n_{Na}=\dfrac{4,6}{23}=0,2mol\)
\(4Na+O_2\rightarrow\left(t^o\right)2Na_2O\)
0,2 0,05 ( mol )
\(V_{O_2}=0,05.24,79=1,2395l\)
4.\(n_{Cu}=\dfrac{1,6}{64}=0,025mol\)
\(2Cu+O_2\rightarrow\left(t^o\right)2CuO\)
0,025 0,0125 ( mol )
\(V_{O_2}=0,0125.24,79=0,309875l\)
n C3H8=0,1 mol
C3H8+5O2-to>3CO2+4H2O
0,1-----0,5------------0,3-----0,4 mol
=>VCO2=0,3.22,4=6,72 l
=>VO2=0,5.22,4=11,2l
\(2CO+O_2\rightarrow\left(t^o\right)2CO_2\\ n_{CO}=\dfrac{2,479}{24,79}=0,1\left(mol\right)\\ n_{O_2}=\dfrac{0,1}{2}=0,05\left(mol\right)\\ V_{O_2\left(đkc\right)}=0,05.24,79=1,2395\left(l\right)\)
mình cùng ko bt thế này đúng ko
2CO+O2→(to)2CO2nCO=2,47924,79=0,1(mol)nO2=0,12=0,05(mol)VO2(đkc)=0,05.24,79=1,2395(l)
a) 2Na + H2SO4 --> Na2SO4 + H2
b) \(n_{Na}=\dfrac{4,6}{23}=0,2\left(mol\right)\)
PTHH: 2Na + H2SO4 --> Na2SO4 + H2
_____0,2------>0,1-------------------->0,1
=> VH2 = 0,1.22,4 = 2,24 (l)
c) mH2SO4 = 0,1.98 = 9,8(g)
a, PT: \(S+O_2\underrightarrow{t^o}SO_2\)
\(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
b, Ta có: \(n_{SO_2}=\dfrac{2,479}{24,79}=0,1\left(mol\right)\)
Theo PT: \(n_S=n_{SO_2}=0,1\left(mol\right)\)
⇒ mP = 6,3 - mS = 6,3 - 0,1.32 = 3,1 (g)
\(\Rightarrow n_P=\dfrac{3,1}{31}=0,1\left(mol\right)\)
Theo PT: \(n_{O_2}=n_S+\dfrac{5}{4}n_P=0,225\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,225.24,79=5,57775\left(l\right)\)
c, Theo PT: \(n_{P_2O_5}=\dfrac{1}{2}n_P=0,05\left(mol\right)\)
\(\Rightarrow m_{P_2O_5}=0,05.142=7,1\left(g\right)\)
Bài 12:
\(n_{Na}=\dfrac{4,6}{23}=0,2\left(mol\right)\)
\(n_{Ca}=\dfrac{8}{40}=0,2\left(mol\right)\)
PTHH: 2Na + 2H2O --> 2NaOH + H2
0,2----------------------->0,1
Ca + 2H2O --> Ca(OH)2 + H2
0,2---------------------->0,2
=> VH2 = (0,1 + 0,2).24,79 = 7,437(l)
nNa = 4,6 : 23 = 0,2 (mol)
2Na + 2H2O - > 2NaOH + H2
0,2 0,1
nCa = 8 : 40 = 0,2 (mol)
Ca + 2H2O -- > Ca(OH)2 + H2
0,2 0,2
nH2 = 0,1 + 0,2 = 0,3 (mol)
VH2 = 0,3 . 24,79 = 7,437 (l)
\(CH_4+2O_2\rightarrow\left(t^o\right)CO_2+2H_2O\\ n_{CH_4}=\dfrac{14,874}{22,79}=0,6\left(mol\right)\\ \Rightarrow n_{CO_2}=n_{CH_4}=0,6\left(mol\right)\\ n_{O_2}=n_{H_2O}=2.0,6=1,2\left(mol\right)\\ V_{O_2\left(đkc\right)}=1,2.24,79=29,748\left(l\right)\\ V_{kk\left(đkc\right)}=29,748.5=148,74\left(l\right)\\ V_{CO_2\left(đkc\right)}=0,6.24,79=14,874\left(l\right)\\ m_{CO_2}=44.0,6=26,4\left(g\right)\\ m_{H_2O}=1,2.18=21,6\left(g\right)\\ V_{H_2O}=\dfrac{21,6}{1}=21,6\left(ml\right)\)
Đề cho đkc nên anh tính theo đkc nhé!
\(pthh:CH_4+2O_2\overset{t^o}{--->}CO_2\uparrow+2H_2O\)
Ta có: \(n_{CH_4}=\dfrac{14,874}{22,4}=\dfrac{7437}{11200}\left(mol\right)\)
Theo pt: \(n_{O_2}=n_{H_2O}=2.n_{CH_4}=2.\dfrac{7437}{11200}\approx1,328\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}V_{O_2}=1,328.22,4=29,7472\left(lít\right)\\m_{H_2O}=1,328.18=23,904\left(g\right)\end{matrix}\right.\)
Theo pt: \(n_{CO_2}=n_{CH_4}=\dfrac{7437}{11200}\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}V_{CO_2}=\dfrac{7437}{11200}.22,4=14,874\left(lít\right)\\m_{CO_2}=\dfrac{7437}{11200}.44\approx29,22\left(g\right)\end{matrix}\right.\)
a) \(n_{H_2}=\dfrac{2,479}{24,79}=0,1\left(mol\right)\)
PTHH: 2H2 + O2 --to--> 2H2O
0,1->0,05
=> VO2 = 0,05.24,79 = 1,2395 (l)
b) \(n_S=\dfrac{12}{32}=0,375\left(mol\right)\)
PTHH: S + O2 --to--> SO2
0,375->0,375
=> VO2 = 0,375.24,79 = 9,29625(l)
c) \(n_{Na}=\dfrac{69}{23}=3\left(mol\right)\)
PTHH: 4Na + O2 --to--> 2Na2O
3-->0,75
=> VO2 = 0,75.24,79 = 18,5925 (l)