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a) C + O2 --to--> CO2
b) \(n_C=\dfrac{24}{12}=2\left(mol\right)\)
=> nCO2 =2 (mol)
=> mCO2 = 2.44 = 88(g)
c)
nO2 = 2(mol)
=> VO2 = 2.22,4 = 44,8 (l)
=> Vkk = 44,8.5=224(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
\(b.C+O_2\xrightarrow[t^0]{}CO_2\\ \Rightarrow n_{CO_2}=n_{O_2}=2mol\\ m_{CO_2}=2.44=88\left(g\right)\)
\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\
pthh:2Mg+O_2\underrightarrow{t^o}2MgO\)
0,2 0,1 0,2
\(V_{O_2}=0,1.22,4=2,24L\\
m_{MgO}=0,2.40=8g\)
\(n_C=\dfrac{3}{12}=0,25\left(mol\right)\)
\(pthh:C+O_2\underrightarrow{t^o}CO_2\)
\(LTL:0,25>0,1\)
=> C không cháy hết
\(1,2H_2+O_2\underrightarrow{t}2H_2O\)
\(2Mg+O_2\underrightarrow{t}2MgO\)
\(2Cu+O_2\underrightarrow{t}2CuO\)
\(S+O_2\underrightarrow{t}SO_2\)
\(4Al+3O_2\underrightarrow{t}2Al_2O_3\)
\(C+O_2\underrightarrow{t}CO_2\)
\(4P+5O_2\underrightarrow{t}2P_2O_5\)
\(2,PTHH:C+O_2\underrightarrow{t}CO_2\)
\(a,n_{O_2}=0,2\left(mol\right)\Rightarrow n_{CO_2}=0,2\left(mol\right)\Rightarrow m_{CO_2}=8,8\left(g\right)\)
\(b,n_C=0,3\left(mol\right)\Rightarrow n_{CO_2}=0,3\left(mol\right)\Rightarrow m_{CO_2}=13,2\left(g\right)\)
c, Vì\(\frac{0,3}{1}>\frac{0,2}{1}\)nên C phản ửng dư, O2 phản ứng hết, Bài toán tính theo O2
\(n_{O_2}=0,2\left(mol\right)\Rightarrow n_{CO_2}=0,2\left(mol\right)\Rightarrow m_{CO_2}=8,8\left(g\right)\)
\(3,PTHH:CH_4+2O_2\underrightarrow{t}CO_2+2H_2O\)
\(C_2H_2+\frac{5}{2}O_2\underrightarrow{t}2CO_2+H_2O\)
\(C_2H_6O+3O_2\underrightarrow{t}2CO_2+3H_2O\)
\(4,a,PTHH:4P+5O_2\underrightarrow{t}2P_2O_5\)
\(n_P=1,5\left(mol\right)\Rightarrow n_{O_2}=1,2\left(mol\right)\Rightarrow m_{O_2}=38,4\left(g\right)\)
\(b,PTHH:C+O_2\underrightarrow{t}CO_2\)
\(n_C=2,5\left(mol\right)\Rightarrow n_{O_2}=2,5\left(mol\right)\Rightarrow m_{O_2}=80\left(g\right)\)
\(c,PTHH:4Al+3O_2\underrightarrow{t}2Al_2O_3\)
\(n_{Al}=2,5\left(mol\right)\Rightarrow n_{O_2}=1,875\left(mol\right)\Rightarrow m_{O_2}=60\left(g\right)\)
\(d,PTHH:2H_2+O_2\underrightarrow{t}2H_2O\)
\(TH_1:\left(đktc\right)n_{H_2}=1,5\left(mol\right)\Rightarrow n_{O_2}=0,75\left(mol\right)\Rightarrow m_{O_2}=24\left(g\right)\)
\(TH_2:\left(đkt\right)n_{H_2}=1,4\left(mol\right)\Rightarrow n_{O_2}=0,7\left(mol\right)\Rightarrow m_{O_2}=22,4\left(g\right)\)
\(5,PTHH:S+O_2\underrightarrow{t}SO_2\)
\(n_{O_2}=0,46875\left(mol\right)\)
\(n_{SO_2}=0,3\left(mol\right)\)
Vì\(0,46875>0,3\left(n_{O_2}>n_{SO_2}\right)\)nên S phản ứng hết, bài toán tính theo S.
\(a,\Rightarrow n_S=n_{SO_2}=0,3\left(mol\right)\Rightarrow m_S=9,6\left(g\right)\)
\(n_{O_2}\left(dư\right)=0,16875\left(mol\right)\Rightarrow m_{O_2}\left(dư\right)=5,4\left(g\right)\)
\(6,a,PTHH:C+O_2\underrightarrow{t}CO_2\)
\(n_{O_2}=1,5\left(mol\right)\Rightarrow n_C=1,5\left(mol\right)\Rightarrow m_C=18\left(g\right)\)
\(b,PTHH:2H_2+O_2\underrightarrow{t}2H_2O\)
\(n_{O_2}=1,5\left(mol\right)\Rightarrow n_{H_2}=0,75\left(mol\right)\Rightarrow m_{H_2}=1,5\left(g\right)\)
\(c,PTHH:S+O_2\underrightarrow{t}SO_2\)
\(n_{O_2}=1,5\left(mol\right)\Rightarrow n_S=1,5\left(mol\right)\Rightarrow m_S=48\left(g\right)\)
\(d,PTHH:4P+5O_2\underrightarrow{t}2P_2O_5\)
\(n_{O_2}=1,5\left(mol\right)\Rightarrow n_P=1,2\left(mol\right)\Rightarrow m_P=37,2\left(g\right)\)
\(7,n_{O_2}=5\left(mol\right)\Rightarrow V_{O_2}=112\left(l\right)\left(đktc\right)\);\(V_{O_2}=120\left(l\right)\left(đkt\right)\)
\(8,PTHH:C+O_2\underrightarrow{t}CO_2\)
\(m_C=0,96\left(kg\right)\Rightarrow n_C=0,08\left(kmol\right)=80\left(mol\right)\Rightarrow n_{O_2}=80\left(mol\right)\Rightarrow V_{O_2}=1792\left(l\right)\)
\(9,n_p=0,2\left(mol\right);n_{O_2}=0,3\left(mol\right)\)
\(PTHH:4P+5O_2\underrightarrow{t}2P_2O_5\)
Vì\(\frac{0,2}{4}< \frac{0,3}{5}\)nên P hết O2 dư, bài toán tính theo P.
\(a,n_{O_2}\left(dư\right)=0,05\left(mol\right)\Rightarrow m_{O_2}\left(dư\right)=1,6\left(g\right)\)
\(b,n_{P_2O_5}=0,1\left(mol\right)\Rightarrow m_{P_2O_5}=14,2\left(g\right)\)
a) PT Chữ: Cacbon + khí oxi ---to----> Khí cacbonic
b) Theo ĐLBTKL, ta có:
mCacbon + m(khí oxi) = m(khí cacbonic)
<=>m(khí cacbonic)= 12+32=44(g)
c) C + O2 -to-> CO2
nC=4,8/12=0,4(mol) => nO2=nCO2=nC=0,4(mol)
=>mO2=0,4.32= 12,8(g)
mCO2=44.0,4= 17,6(g)
a) PTHH: \(C_2H_4+3O_2\xrightarrow[]{t^o}2CO_2+2H_2O\)
b) CT: \(m_{C_2H_4}+m_{ O_2}=m_{CO_2}+m_{H_2O}\)
c) áp dụng định luật bảo toàn khối lượng, ta có:
\(m_{C_2H_4}+m_{ O_2}=m_{CO_2}+m_{H_2O}\)
\(28+m_{O_2}=88+36\)
\(\Rightarrow m_{O_2}=\left(88+36\right)-28=96\left(g\right)\)
vậy khối lượng khí oxi đã phản ứng là \(96g\)
Gọi nC = a (mol); nS = b (mol)
12a + 32b = 12 (1)
PTHH:
C + O2 -> (t°) CO2
a ---> a ---> a
S + O2 -> (t°) SO2
b ---> b ---> b
44a + 64b = 28 (2)
Từ (1)(2) => a = 0,2 (mol); b = 0,3 (mol)
nO2 = 0,2 + 0,3 = 0,5 (mol)
VO2 = 0,5 . 22,4 = 11,2 (l)
\(n_C=\dfrac{6}{12}=0,5\left(mol\right)\)
PTHH: C + O2 --to--> CO2
0,5->0,5
=> mO2 = 0,5.32 = 16 (g)
\(n_{CO_2}=\dfrac{44.8}{22.4}=2\left(mol\right)\)
\(C+O_2\underrightarrow{^{^{t^0}}}CO_2\)
\(2..............2\)
\(m_C=2\cdot12=24\left(g\right)\)
\(\Rightarrow A\)