? + ? -> ZnSO4 + H2
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a, PT: \(2C_2H_2+5O_2\underrightarrow{t^o}4CO_2+2H_2O\)
Ta có: \(n_{C_2H_2}=\dfrac{13}{26}=0,5\left(mol\right)\)
Theo PT: \(n_{CO_2}=n_{C_2H_2}=1\left(mol\right)\Rightarrow m_{CO_2}=1.44=44\left(g\right)\)
\(n_{H_2O}=n_{C_2H_2}=0,5\left(mol\right)\Rightarrow m_{H_2O}=0,5.18=9\left(g\right)\)
b, Theo PT: \(n_{O_2}=\dfrac{5}{2}n_{C_2H_2}=1,25\left(mol\right)\Rightarrow V_{O_2}=1,25.22,4=28\left(l\right)\)
\(\Rightarrow V_{kk}=\dfrac{V_{O_2}}{20\%}=140\left(l\right)\)
c, PT: \(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3+H_2O\)
Theo PT: \(n_{CaCO_3}=n_{CO_2}=1\left(mol\right)\Rightarrow m_{CaCO_3}=1.100=100\left(g\right)\)
n C2H2 = (mol)
2C2H2 + 3O2 4CO2 + 2H2O
0,5 --> 0,75 --> 1 --> 0,5
a. mCO2 = n.M = 1. 44 = 44 (gam)
mH2O = n.M = 0,5 . 18 = 9 (gam)
b. V(O2) = n.22,4 = 0,75 . 22,4 = 16,8 (lít)
Ta có Oxi chiếm 20% thể tích không khí
=> V (Không khí) = (lít)
c. CO2 + Ca(OH)2 CaCO3 + H2O
1 --> 1
m CaCO3 = n.M = 1.100 = 100 (gam)
a, PT: \(MgCO_3+2HCl\rightarrow MgCl_2+CO_2+H_2O\)
\(MgO+2HCl\rightarrow MgCl_2+H_2O\)
b, Ta có: \(n_{CO_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
Theo PT: \(n_{MgCO_3}=n_{CO_2}=0,1\left(mol\right)\)
\(\Rightarrow m_{MgCO_3}=0,1.84=8,4\left(g\right)\)
\(\Rightarrow m_{MgO}=10,4-m_{MgCO_3}=2\left(g\right)\)
Ta có: \(\%m_C=\dfrac{12.13}{12.13+1.21+14+16.3}.100\%\approx65,272\%\)
Phần trăm khối lượng nguyên tử cacbon có trong phân tử Salbutamol là
C%
\(MnO_2+4HCl_đ\underrightarrow{t^o}MnCl_2+Cl_2+2H_2O\)
\(Cl_2+H_2\underrightarrow{t^o}2HCl\)
\(HCl+NaOH\rightarrow NaCl+H_2O\)
\(2NaCl+2H_2O\xrightarrow[cmn]{đpdd}2NaOH+Cl_2+H_2\)
MnO2 + 4HCl MnCl2 + Cl2 + 2H2O
Cl2 + H2 2HCl
HCl + NaOH NaCl + H2O
2NaCl + 2H2O 2NaOH + H2 + Cl2
PT: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
Ta có: \(n_{Fe}=\dfrac{6,72}{56}=0,12\left(mol\right)\)
a, \(n_{H_2}=n_{Fe}=0,12\left(mol\right)\Rightarrow V_{H_2}=0,12.24,79=2,9748\left(l\right)\)
b, \(n_{FeSO_4}=n_{Fe}=0,12\left(mol\right)\Rightarrow m_{FeSO_4}=0,12.152=18,24\left(g\right)\)
c, PT: \(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
\(n_{Fe}=\dfrac{2}{3}n_{H_2}=0,08\left(mol\right)\Rightarrow m_{Fe}=0,08.56=4,48\left(g\right)\)
a, PT: \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
Ta có: \(n_{Br_2}=\dfrac{4}{160}=0,025\left(mol\right)\)
Theo PT: \(n_{C_2H_4}=n_{Br_2}=0,025\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,025.22,4}{2,8}.100\%=20\%\\\%V_{CH_4}=100-20=80\%\end{matrix}\right.\)
b, Ta có: \(n_{CH_4}=\dfrac{2,8.80\%}{22,4}=0,1\left(mol\right)\)
PT: \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)
\(C_2H_4+3O_2\underrightarrow{t^o}2CO_2+2H_2O\)
Theo PT: \(n_{CO_2}=n_{CH_4}+2n_{C_2H_4}=0,15\left(mol\right)\)
PT: \(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_{3\downarrow}+H_2O\)
Theo PT: \(n_{CaCO_3}=n_{CO_2}=0,15\left(mol\right)\)
\(\Rightarrow a=m_{CaCO_3}=0,15.100=15\left(g\right)\)
a, PT:
Ta có:
Theo PT:
b, Ta có:
PT:
Theo PT:
PT:
Theo PT:
(1) \(2CH_4\xrightarrow[lln]{1500^oC}C_2H_2+3H_2\)
(2) \(C_2H_2+H_2\underrightarrow{t^o,Pd}C_2H_4\)
(3) \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
(4) \(CH_4+Cl_2\underrightarrow{as}CH_3Cl+HCl\)
(5) \(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)
(6) \(C_2H_2+2H_2\underrightarrow{t^o,Ni}C_2H_6\)
(7) \(C_2H_4+H_2\underrightarrow{t^o,Ni}C_2H_6\)
\(Zn+H_2SO_4->ZnSO_4+H_2\)
phản ứng thế
\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\) - Pư thế