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\(2CuO+C\xrightarrow[]{t^0}2Cu+CO_2\\ 2Fe_2O_3+3C\xrightarrow[]{t^0}4Fe+3CO_2\\ 2Al_2O_3+3C\xrightarrow[]{t^0}4Al+3CO_2\\ CO_2+CaO\rightarrow CaCO_3\\ CaO+H_2O\rightarrow Ca\left(OH\right)_2\\ Ca\left(OH\right)_2+2Al+2H_2O\rightarrow Ca\left(AlO_2\right)_2+3H_2\\ CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\\ Fe+HCl\rightarrow FeCl_2+H_2\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ Ca\left(OH\right)_2+2HCl\rightarrow CaCl_2+2H_2O\\ Ca\left(AlO_2\right)_2+2HCl+2H_2O\rightarrow2Al\left(OH\right)_3+CaCl_2\\ Al\left(OH\right)_3+3HCl\rightarrow AlCl_3+3H_2O\\ 2Cu+O_2\xrightarrow[t^0]{}2CuO\)
\(a)n_{HCl}=\dfrac{47,45}{36,5}=1,3mol\\ n_{CuO}:n_{Fe_2O_3}=1:1\\ \Rightarrow n_{CuO}=n_{Fe_2O_3}\\ 80n_{CuO}+160n_{Fe_2O_3}=24\\ \Rightarrow80n_{CuO}+160n_{CuO}=24\\ \Rightarrow n_{CuO}=n_{Fe_2O_3}=0,1mol\\ CuO+2HCl\rightarrow CuCl_2+H_2O\\ 0,1.......0,2.........0,1..........0,1\\ Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\\ 0,1..........0,6.........0,2..........0,3\\ n_{HCl.pứ}=0,2+0,6=0,8< n_{HCl}\left(1,3\right)\)
Vậy hh X tan hết
\(b)m_{CuCl_2}=0,1.135=13,5g\\
m_{FeCl_3}=0,2.162,5=32,5g\\
c)n_{HCl.dư}=1,3-0,8=0,5mol\\
ZnO+2HCl\rightarrow ZnCl_2+H_2\\
n_{Zn}=\dfrac{0,5}{2}=0,25mol\\
m_{ZnO}=0,25.81=20,25g\)
\(a)n_C=\dfrac{30}{12}=2,5mol\\ n_P=\dfrac{37,2}{31}=1,2mol\\ n_{O_2}=\dfrac{80}{22,4}=\dfrac{25}{7}mol\\ C+O_2\xrightarrow[]{t^0}CO_2\\ 4P+5O_2\xrightarrow[]{t^0}2P_2O_5\\ n_{O_2.cần,.dùng}=2,5+1,2\cdot\dfrac{5}{4}=4mol< n_{O_2}\left(\dfrac{25}{7}\right)\)
Vậy hh Y không cháy hết
\(b)2H_2O\xrightarrow[điện]{phân}2H_2+O_2\\ n_{H_2O}=4.2=8mol\\ m_{H_2O}=8.16=128g\)
Câu 1
\(a)PTHH:Mg+2HCl\rightarrow MgCl_2+H_2\\
b)200ml=0,2l\\
n_{HCl}=0,2.1=0,2mol\\
n_{H_2}=n_{MgCl_2}=\dfrac{1}{2}n_{H_2}=\dfrac{1}{2}\cdot0,2=0,1mol\\
V_{H_2}=0,1.24,79=2,479l\\
c)C_{M_{MgCl_2}}=\dfrac{0,1}{0,2}=0,5M\)
\(n_{BaCl_2}=\dfrac{200.20,8\%}{208}=0,2\left(mol\right)\\ PTHH:BaCl_2+H_2SO_4\rightarrow BaSO_4+2HCl\\ n_{BaSO_4}=n_{H_2SO_4}=n_{BaCl_2}=0,2\left(mol\right)\\ a,m_{kt}=m_{BaSO_4}=233.0,2=46,6\left(g\right)\\ b,C\%_{ddH_2SO_4}=\dfrac{0,2.98}{200}.100\%=9,8\%\)
\(a,Zn+2HCl\rightarrow ZnCl_2+H_2\\ n_{ZnCl_2}=n_{H_2}=n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\\ b,m_{ZnCl_2}=136.0,1=13,6\left(g\right);V_{H_2\left(đktc\right)}=0,1.22,4=2,24\left(l\right)\\ c,Vì:\dfrac{0,1}{1}>\dfrac{0,2}{1}\Rightarrow n_{Zn\left(TT\right)}=0,1\left(mol\right);n_{Zn\left(LT\right)}=n_{ZnCl_2}=0,2\left(mol\right)\\ H=\dfrac{0,1}{0,2}.100\%=50\%\)
\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right);n_{ZnCl_2}=\dfrac{20,6}{136}\approx0,15147\left(mol\right)\\ PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ Vì:\dfrac{0,2}{1}>\dfrac{0,15147}{1}\\ Có:m_{ZnCl_2\left(LT\right)}=0,2.136=27,2\left(g\right)\\ H=\dfrac{20,6}{27,2}.100\%\approx75,735\%\)
a)
\(PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
b)
\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
Theo PT:\(n_{H_2}=n_{Zn}=0,2mol\)
\(\Rightarrow V_{H_2}=0,2.24,79=4,958l\)
a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
b, \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
Theo PT: \(n_{H_2}=n_{Zn}=0,2\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,2.24,79=4,958\left(l\right)\)
\(39,6g?\\ CTPT\left(A\right):C_xH_y\\ n_C=n_{CO_2}=\dfrac{39,6}{44}=0,9mol\\ m_C=0,9.12=10,8g\\ m_H=11,7-10,8=0,9g\\ M_A=2,689.29=77,981g/mol\)
Ta có tỉ lệ
\(\dfrac{12x}{10,8}=\dfrac{y}{0,9}=\dfrac{77,981}{11,7}\\ \Rightarrow x\approx6;y\approx6\\ \Rightarrow CTPT\left(A\right):C_6H_6\\ 2C_6H_6+15O_2\rightarrow12CO_2+6H_2O\\ n_{H_2O}=\dfrac{0,9.6}{12}=0,45mol\\ m=m_{H_2O}=0,45.18=8,1g\)
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