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\(a,m_C=48\left(g\right)\rightarrow n_C=\dfrac{m_C}{M_C}=\dfrac{48}{12}=4\left(mol\right)\)
\(V_{O_2}=44,8\left(l\right)\rightarrow n_{O_2}=\dfrac{V_{O_2}}{22,4}=\dfrac{44,8}{22,4}=2\left(mol\right)\)
\(PTHH:C+O_2\underrightarrow{t^o}CO_2\)
\(pt:\) \(1mol\) \(1mol\)
\(đb:\) \(4mol\) \(2mol\)
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
\(b.C+O_2\xrightarrow[t^0]{}CO_2\\ \Rightarrow n_{CO_2}=n_{O_2}=2mol\\ m_{CO_2}=2.44=88\left(g\right)\)
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
$n_{Cl_2\ pư} = n_{Mg} = 0,1(mol) \Rightarrow m_{Cl_2\ dư} = (0,4 - 0,1).71 = 21,3(gam)$
$n_{MgCl_2}= n_{Mg} = 0,1(mol) \Rightarrow m_{MgCl_2} = 0,1.95 = 9,5(gam)$
\(n_P=\dfrac{6,2}{31}=0,2\left(mol\right)\\ n_{O_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\\ PTHH:4P+5O_2\underrightarrow{t^o}2P_2O_5\\ LTL:\dfrac{0,2}{4}< \dfrac{0,4}{5}\Rightarrow O_2dư\)
\(n_{O_2\left(pư\right)}=\dfrac{5}{4}n_P=\dfrac{5}{4}.0,2=0,25\left(mol\right)\\ n_{O_2\left(dư\right)}=0,4-0,25=0,15\left(mol\right)\)
\(n_{P_2O_5\left(lt\right)}=\dfrac{1}{2}n_P=\dfrac{1}{2}.0,2=0,1\left(mol\right)\\ m_{P_2O_5\left(lt\right)}=0,1.142=14,2\left(g\right)\\ m_{P_2O_5\left(tt\right)}=0,1.142.80\%=11,36\left(g\right)\)
$a) CH_4 + 2O_2 \xrightarrow{t^o} CO_2 + 2H_2O$
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 :
$n_{Na_2CO_3} = n_{CO_2} = \dfrac{1}{2}n_{CH_4} = 0,06(mol)$
$m_{Na_2CO_3} = 0,06.106 = 6,36(gam)$
\(n_P=\dfrac{3,1}{31}=0,1\left(mol\right)\)
\(n_{O_2}=\dfrac{4,48}{22,4}=0,2mol\)
4P + 5O2 \(\underrightarrow{t^o}\) 2P2O5
\(\dfrac{0,1}{4}< \dfrac{0,2}{5}\) => O2 dư, Photpho đủ
\(n_{O_2}=0,2-0,04=0,16\left(mol\right)\)
\(m_{P_2O_5}=\) 0,05 . 142 = 7,1 ( g )
a, PTHH: S + O2 -> (t°) SO2
b, nS = 6,4/32 = 0,2 (mol)
nO2 = 6,72/22,4 = 0,3 (mol)
LTL: 0,2 < 0,3 => O2 dư
nO2 (pư) = nSO2 = nS = 0,2 (mol)
mO2 (dư) = (0,3 - 0,2) . 32 = 3,2 (g)
c, mSO2 = 64 . 0,2 = 12,8 (g)
a, \(S+O_2\underrightarrow{t^o}SO_2\)
\(nS=\dfrac{6,4}{32}=0,2\left(mol\right)\)
\(nO_2=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
\(\dfrac{0,2}{1}< \dfrac{0,3}{1}\) => oxi dư
\(nO_{2\left(dư\right)}=0,1\left(mol\right)\)
\(mO_{2\left(dư\right)}=0,1.32=3,2\left(g\right)\)
\(nSO_2=nS=0,2\left(mol\right)\)
\(mSO_2=0,2.64=12,8\left(g\right)\)
4P + 5O2 -----to---> 2P2O5
0,4---0,5-----------> 0,2 (mol)
+ n P = 12,4 / 31 = 0,4 (mol)
+nO2 = 13,44 / 22,4 = 0,6 (mol)
Vì nP/4 = 0,1 < n O2 /5 = 0,12
=> Oxi còn thừa sau phản ứng .
mO2 dư = (0,6 - 0,5 ) . 32 = 3,2 (g)
b. chất tạo thành : P2O5
mP2O5 = 0,2 . ( 2.31 + 16 . 5 ) = 28,4 (g)
\(n_P=\dfrac{12,4}{31}=0,4\left(mol\right)\\
n_{O_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\\
pthh:4P+5O_2\underrightarrow{t^o}2P_2O_5\\
LTL:\dfrac{0,4}{4}< \dfrac{0,6}{5}\)
=> Oxi dư
\(n_{O_2\left(p\text{ư}\right)}=\dfrac{5}{4}n_P=0,5\left(mol\right)\\
m_{O_2\left(d\right)}=\left(0,6-0,5\right).32=3,2\left(g\right)\\
n_{P_2O_5}=\dfrac{1}{2}n_P=0,2\left(mol\right)\\
m_{P_2O_5}=0,2.142=28,4\left(g\right)\)
\(n_S=\dfrac{3,2}{32}=0,1\left(mol\right)\\
n_{O_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\\
pthh:S+O_2\underrightarrow{t^o}SO_2\\
LTL:\dfrac{0,1}{1}>\dfrac{0,05}{1}\)
=> S dư
\(n_{S\left(P\text{Ư}\right)}=n_{SO_2}=n_{O_2}=0,05\left(mol\right)\\
m_S=\left(0,1-0,05\right).32=1,6\left(g\right)\\
V_{SO_2}=0,05.22,4=1,12\left(l\right)\)
a) \(n_C=\frac{12}{12}=1\left(mol\right)\)
\(n_{O2}=\frac{64}{32}=2\left(mol\right)\)
________\(C+O_2\underrightarrow{t^O}CO_2\)
Ban đầu: 1____2
Phản ứng: 1___________1(mol)
Dư: 1
Lập tỉ lệ: \(\frac{1}{1}< \frac{2}{1}\left(1< 2\right)\)
b) O2 dư
\(m_{O2\left(dư\right)}=1.32=32\left(g\right)\)
c) \(V_{CO2}=1.44=44\left(g\right)\)
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