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\(n_{C_2H_4}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\\ PTHH:C_2H_4+3O_2\underrightarrow{t^o}2CO_2+2H_2O\)
0,25 0,5
\(\rightarrow m_{H_2O}=0,5.18=9\left(g\right)\)
a, \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)
\(C_2H_4+3O_2\underrightarrow{t^o}2CO_2+2H_2O\)
\(2C_2H_2+5O_2\underrightarrow{t^o}4CO_2+2H_2O\)
b, Sửa đề: 17,9 (l) → 17,92 (l)
Ta có: \(n_{CO_2}=\dfrac{17,92}{22,4}=0,8\left(mol\right)=n_C\)
\(n_{H_2O}=\dfrac{18}{18}=1\left(mol\right)\Rightarrow n_H=1.2=2\left(mol\right)\)
⇒ mA = mC + mH = 0,8.12 + 2.1 = 11,6 (g)
Theo ĐLBT KL, có: mA + mO2 = mCO2 + mH2O
⇒ mO2 = 0,8.44 + 18 - 11,6 = 41,6 (g)
\(\Rightarrow n_{O_2}=\dfrac{41,6}{32}=1,3\left(mol\right)\Rightarrow V_{O_2}=1,3.22,4=29,12\left(l\right)\)
Bài 1:
PTHH: \(2C_4H_{10}+13O_2\xrightarrow[]{t^o}8CO_2+10H_2O\)
Ta có: \(n_{C_4H_{10}}=\dfrac{11,6}{58}=0,2\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{CO_2}=0,8\left(mol\right)\\n_{H_2O}=1\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{CO_2}=0,8\cdot44=35,2\left(g\right)\\m_{H_2O}=1\cdot18=18\left(g\right)\end{matrix}\right.\)
Bài 2:
PTHH: \(CaCO_3\xrightarrow[]{t^o}CaO+CO_2\uparrow\)
Ta có: \(n_{CO_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)=n_{CaO}=n_{CaCO_3\left(p.ứ\right)}\)
\(\Rightarrow\left\{{}\begin{matrix}m_{CaO}=0,5\cdot56=28\left(g\right)\\\%m_{CaCO_3\left(p.ứ\right)}=\dfrac{0,5\cdot100}{100}\cdot100\%=50\%\end{matrix}\right.\)
a) \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
\(n_{HCl}=0,2.1=0,2\left(mol\right)\)
PTHH: Zn + 2HCl --> ZnCl2 + H2
Xét tỉ lệ: \(\dfrac{0,2}{1}>\dfrac{0,2}{2}\) => Zn dư, HCl hết
PTHH: Zn + 2HCl --> ZnCl2 + H2
__________0,2-------------->0,1
=> VH2 = 0,1.22,4 = 2,24(l)
b)
PTHH: 2H2 + O2 --to--> 2H2O
______0,1->0,05
=> mO2 = 0,05.22,4 = 1,12 (l)
\(a,2KHCO_3\rightarrow\left(t^o\right)K_2CO_3+CO_2+H_2O\\ b.n_{K_2CO_3}=\dfrac{27,6}{138}=0,2\left(mol\right)\\ n_{KHCO_3}=2.0,2=0,4\left(mol\right)\\ m_{KHCO_3}=100.0,4=40\left(g\right)\\ c,n_{CO_2}=n_{H_2O}=n_{K_2CO_3}=0,2\left(mol\right)\\ V_{hh\left(CO_2,H_2O\right)}=\left(0,2+0,2\right).22,4=8,96\left(l\right)\)
a, PTHH: 2KHCO3 -to-> K2CO3 + H2O + CO2
b, nK2CO3 = m/M = 27,6/138 = 0,2 (mol)
Theo PTHH: nKHCO3 = 2.nK2CO3 = 2 . 0,2 = 0,4 (mol)
=> mKHCO3 = n.M = 0,4 . 100 = 40 (g)
c, Theo PTHH: nCO2 = nH2O = nK2CO3 = 0,2 (mol)
=> nhh = nCO2 + nH2O = 0,2 + 0,2 = 0,4 (mol)
Ta có: Thể tích khí ở 20oC và 1 atm là đkt
=> Vhh(đkt) = n.24 = 0,4 . 24 = 9,6 (l)
TK:
https://lazi.vn/edu/exercise/452918/dot-chay-16g-chat-a-can-4-48-lit-khi-oxi-o-dktc-thu-duoc-khi-co2-va-hoi-nuoc-theo-ti-le-so-mol-la-1-2-tinh-khoi-luong
a,b,
\(n_{KClO_3}=\dfrac{12,25}{122,5}=0,1\left(mol\right)\)
PTHH: 2KClO3 --to, MnO2--> 2KCl + 3O2
0,1-------------------->0,1------->0,15
\(\rightarrow\left\{{}\begin{matrix}V_{O_2}=0,15.22,4=3,36\left(l\right)\\m_{KCl}=74,5.0,1=7,45\left(g\right)\end{matrix}\right.\)
c, PTHH: 3Fe + 2O2 --to--> Fe3O4
0,225<-0,15------->0,075
=> mFe3O4 = 0,075.232 = 17,4 (g)
a) CH4 + 2O2 --to--> CO2 + 2H2O
b) \(V_{O_2}=\dfrac{56}{5}=11,2\left(l\right)\)
=> \(n_{O_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)
PTHH: CH4 + 2O2 --to--> CO2 + 2H2O
_____0,25<--0,5-------->0,25
=> VCH4 = 0,25.22,4 = 5,6 (l)
c)
mCO2 = 0,25.44 = 11 (g)
Ta có: \(n_{C_4H_{10}}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
PT: \(2C_4H_{10}+13O_2\underrightarrow{t^o}8CO_2+10H_2O\)
a, Theo PT: \(n_{O_2}=\dfrac{13}{2}n_{C_4H_{10}}=1,3\left(mol\right)\)
\(\Rightarrow V_{O_2}=1,3.22,4=29,12\left(l\right)\)
b, Theo PT: \(n_{CO_2}=4n_{C_4H_{10}}=0,8\left(mol\right)\)
\(\Rightarrow m_{CO_2}=0,8.44=35,2\left(g\right)\)
c, PT: \(CO_2+2KOH\rightarrow K_2CO_3+H_2O\)
Theo PT: \(n_{K_2CO_3}=n_{CO_2}=0,8\left(mol\right)\)
\(\Rightarrow m_{K_2CO_3}=0,8.138=110,4\left(g\right)\)
2C4H10 + 13O2 = nhiệt độ => 8CO2 + 10H2O
nC4H10= \(\dfrac{4,48}{22,4}\)= 0,2 (mol)
=> nCO2= 5.nC4H10= 5.0,2 = 1 (mol)
=> mCO2= 1.44=44 (g)
nO2=\(\dfrac{13}{2.n_{C4H10}}\)= \(\dfrac{13}{2}\).0,2= 1,3 (mol)
=> VO2= 1,3 . 22,4= 29,12 (l)