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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)
$n_{CO_2} = 3n_{C_3H_8} = 3.\dfrac{1,344}{22,4} = 0,18(mol)$
$m_{CO_2} = 0,18.44 = 7,92(gam)$
$n_{H_2O} = 4n_{C_3H_8} = 0,24(mol)$
$m_{H_2O} = 0,24.18 = 4,32(gam)$
a, Theo gt ta có: $n_{H_2}=0,15(mol);n_{O_2}=0,05(mol)$
$2H_2+O_2\rightarrow 2H_2O$
Sau phản ứng $H_2$ còn dư. Và dư 0,05.22,4=1,12(l)
b, Ta có: $n_{H_2O}=2.n_{O_2}=0,1(mol)\Rightarrow m_{H_2O}=1,8(g)$
\(n_{SO_2}=\dfrac{V_{SO_2\left(ĐKTC\right)}}{22,4}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
PTHH: \(S+O_2\underrightarrow{t^o}SO_2\)
...........1.........1........1......
...........0,3......0,3......0,3.....
a. \(m_S=n_S\cdot M_S=0,3\cdot32=9,6\left(g\right)\)
b. \(V_{O_2\left(ĐKTC\right)}=n_{O_2}\cdot22,4=0,3\cdot22,4=6,72\left(l\right)\)
\(V_{kk\left(ĐKTC\right)}=V_{O_2\left(ĐKTC\right)}\cdot5=6,72\cdot5=33,6\left(l\right)\)
Theo gt ta có: $n_{H_2}=0,75(mol)$
a, $2H_2+O_2\rightarrow 2H_2O$
Ta có: $n_{O_2}=0,5.n_{H_2}=0,375(mol)\Rightarrow V_{O_2}=8,4(l)\Rightarrow V_{kk}=42(l)$
b, $2KMnO_4\rightarrow K_2MnO_4+MnO_2+O_2$
Ta có: $n_{KMnO_4}=2.n_{O_2}=0,75(mol)\Rightarrow m_{KMnO_4}=118,5(g)$
a)
\(2H_2 + O_2 \xrightarrow{t^o} 2H_2O\\ V_{O_2} = \dfrac{V_{H_2}}{2} = \dfrac{16,8}{2} = 8,4(lít)\\ V_{không\ khí} = \dfrac{8,4}{20\%} = 42(lít)\)
b)
\(n_{O_2} = \dfrac{8,4}{22,4} = 0,375(mol)\\ 2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2\\ n_{KMnO_4} = 2n_{O_2} = 0,75(mol)\\ \Rightarrow m_{KMnO_4} = 0,75.158 = 118,5(gam)\\ 2KClO_3 \xrightarrow{t^o} 2KCl + 3O_2\\ n_{KClO_3} = \dfrac{2}{3}n_{O_2} = 0,25(mol)\\ \Rightarrow m_{KClO_3} = 0,25.122,5 = 30,625(gam)\)
2H2+O2-to>2H2O
0,1----0,05----0,1
n H2=0,1 mol
=>m H2OI=0,1.18=1,8g
=>Vkk=0,05.22,4.5=5,6l
PT: \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)
Ta có: \(n_{CH_4}=\dfrac{30,9875}{24,79}=1,25\left(mol\right)\)
a, \(n_{H_2O}=2n_{CH_4}=2,5\left(mol\right)\) \(\Rightarrow m_{H_2O}=2,5.18=45\left(g\right)\)
b, \(n_{O_2}=2n_{CH_4}=2,5\left(mol\right)\) \(\Rightarrow V_{O_2}=2,5.24,79=61,975\left(l\right)\)
Mà: O2 chiếm 1/5 thể tích không khí.
\(\Rightarrow V_{kk}=5V_{O_2}=309,875\left(l\right)\)
\(n_{H_2}\)=\(\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
PTHH 2H2 +O2----to--->2H2O
0,2....0,1.................0,2
=>\(m_{H_2O}=0,2.18=3,6\left(g\right)\)
=>\(V_{O_2}=0,1.22,4=2,24\left(l\right)\)
=>Vkk=2,24.5=11,2(l)
\(n_{H_2} = \dfrac{4,48}{22,4} = 0,2(mol)\\ 2H_2 + O_2 \xrightarrow{t^o} 2H_2O\\ n_{H_2O} = n_{H_2} =0,2(mol) \Rightarrow m_{H_2O} = 0,2.18 = 3,6(gam)\\ n_{O_2} = \dfrac{1}{2}n_{H_2} = 0,1(mol)\\ \Rightarrow V_{O_2} = 0,1.22,4 = 2,24(lít)\\ \Rightarrow V_{không\ khí} = 5V_{O_2} = 2,24.5 = 11,2(lít) \)
V(O2)= 20%.Vkk=20%. 44,8= 8,92(l) => nO2=0,4(mol)
nC3H8= 1,344/22,4= 0,06(mol)
PTHH: C3H8 + 5 O2 -to-> 3 CO2 + 4 H2O
Ta có: 0,06/1 < 0,4/5
=> O2 dư, C3H8 hết, tính theo nC3H8
=> nO2(p.ứ)= 0,06.5=0,3(mol)=> nO2(dư)=0,4-0,3=0,1(mol)
=> V(O2,dư)=0,1.22,4=2,24(l)
b) nCO2=3.0,06=0,18(mol)
=>mCO2=0,18 . 44=7,92(g)
nH2O=0,06.4=0,24(mol)
=>mH2O=0,24.18=4,32g)
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