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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)
Chúc em học tốt!
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)
$n_{CO_2} = 3n_{C_3H_8} = 3.\dfrac{1,344}{22,4} = 0,18(mol)$
$m_{CO_2} = 0,18.44 = 7,92(gam)$
$n_{H_2O} = 4n_{C_3H_8} = 0,24(mol)$
$m_{H_2O} = 0,24.18 = 4,32(gam)$
\(a,m_C=48\left(g\right)\rightarrow n_C=\dfrac{m_C}{M_C}=\dfrac{48}{12}=4\left(mol\right)\)
\(V_{O_2}=44,8\left(l\right)\rightarrow n_{O_2}=\dfrac{V_{O_2}}{22,4}=\dfrac{44,8}{22,4}=2\left(mol\right)\)
\(PTHH:C+O_2\underrightarrow{t^o}CO_2\)
\(pt:\) \(1mol\) \(1mol\)
\(đb:\) \(4mol\) \(2mol\)
Xét tỉ lệ:
\(\dfrac{n_{C\left(đb\right)}}{n_{C\left(pt\right)}}=\dfrac{4}{1}=4>\dfrac{n_{O_2\left(đb\right)}}{n_{O_2\left(pt\right)}}=\dfrac{2}{1}=2\)
\(\Rightarrow\) \(O_2\) hết, \(C\) dư.
\(b,PTHH:C+O_2\underrightarrow{t^o}CO_2\)
\(pt:\) \(1mol\) \(1mol\)
\(đb:\) \(2mol\) \(2mol\)
\(\Rightarrow m_{CO_2}=n_{CO_2}.M_{CO_2}=2.\left(1.C+2.O\right)=2.\left(1.12+2.16\right)=88\left(g\right)\)
\(a.n_C=\dfrac{48}{12}=4\left(mol\right);n_{O_2}=\dfrac{44,8}{22,4}=2\left(mol\right)\\ C+O_2\xrightarrow[t^0]{}CO_2\)
Theo pt:\(\dfrac{4}{1}>\dfrac{2}{1}\Rightarrow C\) dư, O2 pư hết
\(b.C+O_2\xrightarrow[t^0]{}CO_2\\ \Rightarrow n_{CO_2}=n_{O_2}=2mol\\ m_{CO_2}=2.44=88\left(g\right)\)
\(a,Đặt:n_{CH_4}=a\left(mol\right);n_{C_4H_{10}}=b\left(mol\right)\left(a,b>0\right)\\ PTHH:CH_4+2O_2\rightarrow\left(t^o\right)CO_2+2H_2O\\ 2C_4H_{10}+13O_2\rightarrow\left(t^o\right)8CO_2+10H_2O\\ \Rightarrow\left\{{}\begin{matrix}16a+58b=7,4\\22,4a+22,4.4b=22\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}m_{CH_4}=0,1.16=1,6\left(g\right)\\m_{C_4H_{10}}=0,1.58=5,8\left(g\right)\end{matrix}\right.\\ b,n_{O_2}=2a+\dfrac{13}{2}b=2.0,1+6,5.0,1=0,85\left(mol\right)\\ \Rightarrow V_{O_2\left(đktc\right)}=0,85.22,4=19,04\left(l\right)\)
\(n_{Fe}=\dfrac{12.6}{56}=0.225\left(mol\right)\)
\(n_{O_2}=\dfrac{4.2}{22.4}=0.1875\left(mol\right)\)
\(3Fe+2O_2\underrightarrow{^{^{t^0}}}Fe_3O_4\)
\(3.........2\)
\(0.225......0.1875\)
Lập tỉ lệ : \(\dfrac{0.225}{3}< \dfrac{0.1875}{2}\Rightarrow O_2dư\)
\(m_{O_2\left(dư\right)}=\left(0.1875-0.225\cdot\dfrac{2}{3}\right)\cdot32=1.2\left(g\right)\)
\(m_{Fe_3O_4}=\dfrac{0.225}{3}\cdot232=17.4\left(g\right)\)
nC=4,8/12=0,4(mol)
nO2=6,72/22,4=0,3(mol)
PTHH: C+ O2 -to-> CO2
Ta có: 0,4/1 > 0,3/1
=> C dư, O2 hết, tính theo nO2
=> nCO2=nC(p.ứ)=nO2=0,3(mol)
=>nC(dư)=0,4-0,3=0,1(mol)
=>mC(dư)=0,1.12=1,2(g)
V(CO2,đktc)=V(O2,đktc)=6,72(l) (Số mol tỉ lệ thuận thể tích)
a. PTHH: \(C_3H_8+5O_2\rightarrow^{t^o}3CO_2\uparrow+4H_2O\)
0,6 1,8 2,4 mol
b. \(n_{C_3H_8}=\frac{V}{22,4}=\frac{13,44}{22,4}=0,6mol\)
\(\rightarrow n_{CO_2}=n_{C_3H_8}=\frac{0,6.3}{1}=1,8mol\)
\(m_{CO_2}=n.M=1,8.44=79,2g\)
\(\rightarrow n_{H_2O}=n_{C_3H_8}=\frac{0,6.4}{1}=2,4mol\)
\(m_{H_2O}=n.M=2,4.18=43,2g\)