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Câu 1 :
$n_C = \dfrac{4,8}{12} = 0,4(mol) ; n_{O_2} = \dfrac{7,437}{24,79} = 0,3(mol)$$
$C + O_2 \xrightarrow{t^o} CO_2$
Ta thấy :
$n_C : 1 > n_{O_2} : 1$ nên C dư
$n_{C\ pư} = n_{O_2} = 0,3(mol) \Rightarrow m_{C\ dư} = (0,4 - 0,3).12 = 1,2(gam)$
$\Rightarorw V_{CO_2} = V_{O_2} = 7,437(lít)$
Câu 2 :
$n_{Mg} = \dfrac{2,4}{24} = 0,1(mol)$
$n_{Cl_2} = \dfrac{9,916}{24,79} = 0,4(mol)$
$Mg + Cl_2 \xrightarrow{t^o} MgCl_2$
Ta thấy :
$n_{Mg} : 1 < n_{Cl_2} : 1$ nên $Cl_2$ dư
$n_{Cl_2\ pư} = n_{Mg} = 0,1(mol) \Rightarrow m_{Cl_2\ dư} = (0,4 - 0,1).71 = 21,3(gam)$
$n_{MgCl_2}= n_{Mg} = 0,1(mol) \Rightarrow m_{MgCl_2} = 0,1.95 = 9,5(gam)$
PTHH: \(C+O_2\xrightarrow[]{t^o}CO_2\)
Ta có: \(\left\{{}\begin{matrix}n_C=\dfrac{6,4}{12}=\dfrac{8}{15}\left(mol\right)\\n_{O_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\) C còn dư
\(\Rightarrow\left\{{}\begin{matrix}n_{C\left(dư\right)}=\dfrac{7}{30}\left(mol\right)\\n_{CO_2}=0,3\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{C\left(dư\right)}=\dfrac{7}{30}\cdot12=2,8\left(g\right)\\V_{CO_2}=0,3\cdot22,4=6,72\left(l\right)\end{matrix}\right.\)
nCnC == 4,812=0,4(mol)4,812=0,4(mol)
nO2nO2 == 6,7222,4=0,3(mol)6,7222,4=0,3(mol)
PTHH: C+O2C+O2 to→→to CO2CO2
Do: 0,4>0,30,4>0,3 →→ CC dư
Theo PT: nC(pư)nC(pư) == nCO2nCO2 == nO2nO2 == 0,3(mol)0,3(mol)
nC(dư)nC(dư) == 0,4−0,3=0,1(mol)0,4-0,3=0,1(mol)
mC(dư)mC(dư) == 0,1.12=1,2(g)0,1.12=1,2(g)
VCO2VCO2 == 0,3.22,4=6,72(l)
\(a,m_C=48\left(g\right)\rightarrow n_C=\dfrac{m_C}{M_C}=\dfrac{48}{12}=4\left(mol\right)\)
\(V_{O_2}=44,8\left(l\right)\rightarrow n_{O_2}=\dfrac{V_{O_2}}{22,4}=\dfrac{44,8}{22,4}=2\left(mol\right)\)
\(PTHH:C+O_2\underrightarrow{t^o}CO_2\)
\(pt:\) \(1mol\) \(1mol\)
\(đb:\) \(4mol\) \(2mol\)
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
\(b.C+O_2\xrightarrow[t^0]{}CO_2\\ \Rightarrow n_{CO_2}=n_{O_2}=2mol\\ m_{CO_2}=2.44=88\left(g\right)\)
$a) CH_4 + 2O_2 \xrightarrow{t^o} CO_2 + 2H_2O$
b) $n_{CH_4} = \dfrac{3,92}{22,4} = 0,175(mol)$
$n_{O_2} = \dfrac{3,84}{32} = 0,12(mol)$
Ta thấy : $n_{CH_4} : 1 > n_{O_2} : 2$ nên $CH_4$ dư
$n_{CH_4\ pư} = \dfrac{1}{2}n_{O_2} = 0,06(mol)$
$\Rightarrow m_{CH_4\ dư} = (0,175 - 0,06).16 = 1,84(gam)$
c) $2NaOH + CO_2 \to Na_2CO_3 + H_2O$
Theo PTHH :
$n_{Na_2CO_3} = n_{CO_2} = \dfrac{1}{2}n_{CH_4} = 0,06(mol)$
$m_{Na_2CO_3} = 0,06.106 = 6,36(gam)$
Theo đề bài ta có : \(\left\{{}\begin{matrix}nC=\dfrac{48}{12}=4\left(mol\right)\\nO2=\dfrac{6,72}{22,4}=0,3\left(moL\right)\end{matrix}\right.\)
PTHH :
\(C+O2-^{t0}->CO2\)
0,3mol..0,3mol...0,3mol
Theo PTHH ta có :
\(nC=\dfrac{4}{1}mol>nO2=\dfrac{0,3}{1}mol=>nC\left(d\text{ư}\right)\) ( tính theo nO2 )
=> \(\left\{{}\begin{matrix}mC\left(d\text{ư}\right)=\left(4-0,3\right).12=44,4\left(g\right)\\VCo2\left(\text{đ}ktc\right)=0,3.22,4=6,72\left(l\right)\end{matrix}\right.\)
\(n_{C_2H_2}=\dfrac{6.72}{22.4}=0.3\left(mol\right)\)
\(n_{O_2}=\dfrac{11.2}{22.4}=0.5\left(mol\right)\)
\(C_2H_2+\dfrac{5}{2}O_2\underrightarrow{t^0}2CO_2+H_2O\)
\(Bđ:0.3.......0.5\)
\(Pư:0.2........0.5.........0.4.........0.2\)
\(Kt:0.1..........0..........0.4...........0.2\)
\(V_{CO_2}=0.4\cdot22.4=8.96\left(l\right)\)
\(V_{C_2H_2\left(dư\right)}=0.1\cdot22.4=2.24\left(l\right)\)
a, Ta có: \(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
\(n_{O_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
PT: \(2H_2+O_2\underrightarrow{t^o}2H_2O\)
Xét tỉ lệ: \(\dfrac{0,3}{2}< \dfrac{0,2}{1}\), ta được O2 dư.
Theo PT: \(n_{O_2\left(pư\right)}=\dfrac{1}{2}n_{H_2}=0,15\left(mol\right)\)
\(\Rightarrow n_{O_2\left(dư\right)}=0,05\left(mol\right)\Rightarrow m_{O_2\left(dư\right)}=0,05.32=1,6\left(g\right)\)
b, \(n_{H_2O}=n_{H_2}=0,3\left(mol\right)\)
\(\Rightarrow m_{H_2O}=0,3.18=5,4\left(g\right)\)
c, PT: \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
_______0,3_______________________0,15 (mol)
\(\Rightarrow m_{KMnO_4}=0,3.158=47,4\left(g\right)\)
Bạn tham khảo nhé!
a)
\(n_{Mg} = \dfrac{4,8}{24} = 0,2(mol)\\ n_{O_2} = \dfrac{3,36}{22,4}= 0,15(mol)\\ 2Mg + O_2 \xrightarrow{t^o} 2MgO\)
Ta thấy :
\( \dfrac{n_{Mg}}{2} = 0,1 < n_{O_2} = 0,15 \) nên O2 dư.
\(n_{O_2\ pư} = \dfrac{n_{Mg}}{2} = 0,1(mol)\\ m_{O_2\ dư} = (0,15-0,1).32 = 1,6(gam)\\ V_{O_2\ dư} = (0,15-0,1).22,4 = 1,12(lít)\)
b)
\(n_{MgO} = n_{Mg} = 0,2\ mol\\ \Rightarrow m_{MgO} = 0,2.40 = 8\ gam\)
a, 2Mg + O2 \(\underrightarrow{t^o}\) 2MgO
b, \(n_{Mg}=\dfrac{4,8}{24}=0,2mol\)
\(n_{O_2}=\dfrac{0,2}{2}=0,1mol\)
\(m_{O_2}=0,1.32=3,2g\)
\(V_{O_2}=0,1.22,4=2,24l\)
c, Cách 1:
\(Theo.ĐLBTKL,ta.có:\\ m_{Mg}+m_{O_2}=m_{MgO}\)
\(\Rightarrow m_{MgO}=4,8+3,2=8g\)
Cách 2:
\(n_{MgO}=\dfrac{0,2.2}{2}=0,2mol\)
\(\Rightarrow m_{MgO}=0,2.40=8g\)
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)