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a, \(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)
b, \(n_{KCl}=\dfrac{0,745}{74,5}=0,01\left(mol\right)\)
Theo PT: \(n_{O_2}=\dfrac{3}{2}n_{KCl}=0,015\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,015.24,79=0,37185\left(l\right)\)
\(m_{O_2}=0,015.32=0,48\left(g\right)\)
c, \(n_{KClO_3\left(pư\right)}=n_{KCl}=0,01\left(mol\right)\)
\(\Rightarrow m_{KClO_3\left(pư\right)}=0,01.122,5=1,225\left(g\right)\)
\(\Rightarrow H=\dfrac{1,225}{2,5}.100\%=49\%\)
a) $V_{O_2} = \dfrac{44,8}{5} = 8,96(lít)$
$C_3H_8 + 5O_2 \xrightarrow{t^o} 3CO_2 + 4H_2O$
Ta thấy :
$V_{C_3H_8} : 1 < V_{O_2} :5$ nên Oxi dư
$V_{O_2\ pư} = 5V_{C_3H_8} = 6,72(lít)$
$V_{O_2\ dư} = 8,96 - 6,72 = 2,24(lít)$
b)
$n_{CO_2} = 3n_{C_3H_8} = 3.\dfrac{1,344}{22,4} = 0,18(mol)$
$m_{CO_2} = 0,18.44 = 7,92(gam)$
$n_{H_2O} = 4n_{C_3H_8} = 0,24(mol)$
$m_{H_2O} = 0,24.18 = 4,32(gam)$
\(n_P=\dfrac{7,44}{31}=0,24mol\)
\(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
0,24 0,3 0,12
\(V_{O_2}=0,3\cdot22,4=6,72l\)
\(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)
0,2 0,3
\(m_{KClO_3}=0,2\cdot122,5=24,5g\)
\(m_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
PT: \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
Theo PT: \(n_{O_2}=\dfrac{3}{4}n_{Al}=0,15\left(mol\right)\Rightarrow V_{O_2}=0,15.24,79=3,7185\left(l\right)\)
a) C + O2 --to--> CO2
b) \(n_C=\dfrac{2,4}{12}=0,2\left(mol\right)\)
PTHH: C + O2 --to--> CO2
_____0,2->0,2------>0,2
=> mCO2 = 0,2.44 = 8,8 (g)
c) VO2 = 0,2.22,4 = 4,48(l)
=> Vkk = 4,48.5 = 22,4 (l)
a) \(PTHH:C+O_2\) → \(CO_2\)
bạn xem lại đề nha chỉ làm được mỗi câu a
a) C + O2 --to--> CO2
b) \(n_C=\dfrac{2.4}{12}=0,2\left(mol\right)\)
PTHH: C + O2 --to--> CO2
_____0,2->0,2------>0,2
=> mCO2 = 0,2.44 = 8,8 (g)
c) VO2 = 0,2.22,4 = 4,48(l)
=> Vkk = 4,48.5 = 22,4 (l)
Bài 15:
a) \(n_{O_2}=\dfrac{12,395}{24,79}=0,5\left(mol\right)\)
PTHH: 2H2O --đp--> 2H2 + O2
1<--------------0,5
=> \(H=\dfrac{1.18}{22,5}.100\%=80\%\)
b) \(n_{H_2O}=\dfrac{81}{18}=4,5\left(mol\right)\)
PTHH: 2H2 + O2 --to--> 2H2O
4,5<------------4,5
=> \(V_{H_2\left(lý.thuyết\right)}=4,5.24,79=111,555\left(l\right)\)
=> \(V_{H_2\left(tt\right)}=\dfrac{111,555.100}{90}=123,95\left(l\right)\)
c) \(n_{H_2}=\dfrac{30,9875}{24,79}=1,25\left(mol\right)\)
PTHH: 2H2O --đp--> 2H2 + O2
1,25<-------1,25
=> \(m_{H_2O\left(lý.thuyết\right)}=1,25.18=22,5\left(g\right)\)
=> \(m_{H_2O\left(tt\right)}=\dfrac{22,5.100}{75}=30\left(g\right)\)
\(n_{H_2O}=\dfrac{22,5}{18}=1,25\left(mol\right)\)
\(n_{O_2}=\dfrac{12,395}{24,79}=0,5mol\)
\(2H_2O\rightarrow\left(điện.phân\right)2H_2+O_2\)
1,25 0,5 ( mol ) ( thực tế )
1 0,5 ( mol ) ( lý thuyết )
\(H=\dfrac{1}{1,25}.100=80\%\)
b.\(n_{H_2O}=\dfrac{81}{18}=4,5\left(mol\right)\)
\(2H_2+O_2\rightarrow\left(t^o\right)2H_2O\)
4,5 4,5 ( mol )
\(V_{H_2}=4,5.24,79:90\%=123,95l\)
c.\(n_{H_2}=\dfrac{30,9875}{24,79}=1,25mol\)
\(2H_2O\rightarrow\left(điện.phân\right)2H_2+O_2\)
1,25 1,25 ( mol )
\(m_{H_2O}=1,25.18:75\%=30g\)
a, \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
b, \(n_{Al}=\dfrac{8,1}{27}=0,3\left(mol\right)\)
Theo PT: \(n_{O_2}=\dfrac{3}{4}n_{Al}=0,225\left(mol\right)\Rightarrow V_{O_2}=0,225.22,4=5,04\left(l\right)\)
c, \(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)
Theo PT: \(n_{KClO_3}=\dfrac{2}{3}n_{O_2}=0,15\left(mol\right)\Rightarrow m_{KClO_3}=0,15.122,5=18,375\left(g\right)\)
a, PT: \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
b, Ta có: \(n_{KMnO_4}=\dfrac{3,16}{158}=0,02\left(mol\right)\)
Theo PT: \(n_{O_2}=\dfrac{1}{2}n_{KMnO_4}=0,01\left(mol\right)\)
\(\Rightarrow m_{O_2}=0,01.32=0,32\left(g\right)\)
c, \(V_{O_2}=0,01.24,79=0,2479\left(l\right)\)
\(a)n_{H_2O_2}=\dfrac{50.34\%}{100\%.34}=0,5mol\\ PTHH:2H_2O_2\xrightarrow[]{đpdd}2H_2O+O_2\)
Lí thuyết\(:0,5..............0,5........0,25\)
Thực tế \(:0,5..............0,4.........0,2\)
\(V_{O_2}=0,2.24,79=4,958l\)
b) PỨ đâu cần không khí đâu nhỉ?