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a)
Theo ĐLBTKL: \(m_{Fe\left(bđ\right)}+m_{O_2}=m_X\)
=> \(m_{O_2}=26,4-20=6,4\left(g\right)\)
=> \(n_{O_2}=\dfrac{6,4}{32}=0,2\left(mol\right)\Rightarrow V=0,2.22,4=4,48\left(l\right)\)
b)
PTHH: 3Fe + 2O2 --to--> Fe3O4
0,2------->0,1
=> \(\%m_{Fe_3O_4}=\dfrac{0,1.232}{26,4}.100\%=87,88\%\)
c)
- Nếu dùng KClO3
PTHH: 2KClO3 --to--> 2KCl + 3O2
\(\dfrac{0,4}{3}\)<-----------------0,2
=> \(m_{KClO_3}=\dfrac{0,4}{3}.122,5=\dfrac{49}{3}\left(g\right)\)
- Nếu dùng KMnO4:
PTHH: 2KMnO4 --to--> K2MnO4 + MnO2 + O2
0,4<--------------------------------0,2
=> \(m_{KMnO_4}=0,4.158=63,2\left(g\right)\)
Ta có: \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
Theo PT: \(n_{HCl}=2n_{H_2}=0,2\left(mol\right)\Rightarrow C_{M_{HCl}}=\dfrac{0,2}{0,1}=2\left(M\right)\)
\(n_{Fe}=n_{H_2}=0,1\left(mol\right)\Rightarrow m_{Fe}=0,1.56=5,6\left(g\right)\)
\(\Rightarrow m_{Cu}=20-5,6=14,4\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{5,6}{20}.100\%=28\%\\\%m_{Cu}=72\%\end{matrix}\right.\)
Gọi x, y lần lượt là số mol của Fe và Mg.
Theo đề, ta có: \(56x+24y=13,2\) (*)
Ta có: \(n_{O_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
PTHH:
\(3Fe+2O_2\overset{t^o}{--->}Fe_3O_4\left(1\right)\)
\(2Mg+O_2\overset{t^o}{--->}2MgO\left(2\right)\)
Theo PT(1): \(n_{O_2}=\dfrac{2}{3}.n_{Fe}=\dfrac{2}{3}x\left(mol\right)\)
Theo PT(2): \(n_{O_2}=\dfrac{1}{2}.n_{Mg}=\dfrac{1}{2}y\left(mol\right)\)
\(\Rightarrow\dfrac{2}{3}x+\dfrac{1}{2}y=0,2\) (**)
Từ (*) và (**), ta có HPT:
\(\left\{{}\begin{matrix}56x+24y=13,2\\\dfrac{2}{3}x+\dfrac{1}{2}y=0,2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,15\\y=0,2\end{matrix}\right.\)
\(\Rightarrow m_{Fe}=0,15.56=8,4\left(g\right)\)
\(m_{Mg}=0,2.24=4,8\left(g\right)\)
\(n_{Mg}=\dfrac{2,4}{24}=0,1mol\)
\(n_{Al}=\dfrac{5,4}{27}=0,2mol\)
\(n_{O_2}=\dfrac{26,88:5}{22,4}=0,24mol\)
\(2Mg+O_2\rightarrow\left(t^o\right)2MgO\)
0,1 0,05 0,1 ( mol )
\(4Al+3O_2\rightarrow\left(t^o\right)2Al_2O_3\)
0,2 0,15 0,1 ( mol )
\(n_{O_2\left(td\right)}=0,05+0,15=0,2mol\)
=> Hỗn hợp A cháy hết
\(\left\{{}\begin{matrix}m_{MgO}=0,05.40=2g\\m_{Al_2O_3}=0,1.102=10,2g\end{matrix}\right.\)
Gọi a, b, b là mol Fe, Mg, Cu trong 25,6g X
\(\Rightarrow56a+24b+64b=25,6\left(1\right)\)
Gọi ka, kb, kb là mol Fe, Mg, Cu trong 1 mol X (k > 0)
\(\Rightarrow k\left(a+b+c\right)=1\left(+\right)\)
\(n_{O2}=0,6\left(mol\right)\)
\(3Fe+2O_2\underrightarrow{^{to}}Fe_3O_4\)
\(2Mg+O_2\underrightarrow{^{to}}2MgO\)
\(2Cu+O_2\underrightarrow{^{to}}2CuO\)
\(\Rightarrow k.\left(\frac{2a}{3}+0,5b+0,5b\right)=0,6\left(++\right)\)
Từ (+)(++) \(\Rightarrow k=\frac{1}{a+b+b}=\frac{0,6}{\frac{2a}{3}+0,5b+0,5b}\)
\(\Rightarrow0,6\left(a+2b\right)=\frac{2a}{3}+b\)
\(\Rightarrow\frac{-a}{15}+0,2b=0\left(2\right)\)
Từ (1)(2) \(\Rightarrow\left\{{}\begin{matrix}a=0,3\\b=0,1\end{matrix}\right.\)
\(m_{Fe}=0,3.56=16,8\left(g\right)\)
\(m_{Mg}=0,1.24=2,4\left(g\right)\)
\(m_{Cu}=0,1.64=6,4\left(g\right)\)
$\rm a)n_{kk} = \dfrac{67,2}{22,4} = 3 (mol)$
$\rm \Rightarrow n_{O_2} = 20\%.3 = 0,6 (mol)$
$\rm n_P = \dfrac{24,8}{31} = 0,8 (mol)$
PTHH: \(\rm 4P + 5O_2 \xrightarrow{t^o} 2P_2O_5 \)
Ban đầu: 0,8 0,6
Pư: 0,48<--0,6
Sau pư: 0,32 0 0,24
$\rm \Rightarrow m_{\text{sản phẩm tạo thành}} = m_{P_2O_5(sinh.ra)} = 0,24.142 = 34,08 (g)$
$\m b) m_{hh} = m_{P(dư)} + m_{P_2O_5} = 0,32.31 + 34,08 = 44 (g)$
$\rm \Rightarrow \%m_P = \dfrac{0,32.31}{44} .100\% = 22,545\%$
$\rm \Rightarrow \%m_{P_2O_5} = 100\% - 22,545\% = 77,455\%$
\(n_P=\dfrac{24,8}{31}=0,8\left(mol\right)\)
Thể tích Oxi trong 67,2 lít không khí :
67,2 x 20% = 13,44(l)
\(n_{O_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)
PTHH :
\(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
Trc p/ư: 0,8 0,6 (mol)
p/ư 0,48 0,6 0,24
Sau p/ư: 0,32 0 0,24
=> Sau p/ư P dư
Khối lượng sản phẩm tạo thành :
\(m_{P_2O_5}=0,24.142=34,08\left(g\right)\)
Khối lượng P trong hỗn hợp :
\(m_{P\left(P_2O_5\right)}=0,48.31=14,88\left(g\right)\)
Thành phần % của P :
\(14,88:34,08=43,66\%\)
a)
2CO + O2 --to--> 2CO2
2H2 + O2 --to--> 2H2O
b) \(n_{H_2O}=\dfrac{12,6}{18}=0,7\left(mol\right)\); \(n_{CO_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)
PTHH: 2CO + O2 --to--> 2CO2
0,6<--0,3<------0,6
2H2 + O2 --to--> 2H2O
0,7<--0,35<------0,7
=> \(\left\{{}\begin{matrix}V_{CO}=0,6.22,4=13,44\left(l\right)\\V_{H_2}=0,7.22,4=15,68\left(l\right)\end{matrix}\right.\)
VO2 = (0,3 + 0,35).22,4 = 14,56 (l)
c) \(M_A=\dfrac{0,6.28+0,7.2}{0,6+0,7}=14\left(g/mol\right)\)
=> \(d_{A/O_2}=\dfrac{14}{32}=0,4375\)
\(n_C = a(mol) ; n_S = b(mol)\\ \Rightarrow 12a + 32b = 5,6(1)\\ C + O_2 \xrightarrow{t^o} CO_2\\ S + O_2 \xrightarrow{t^o} SO_2\\ n_{O_2} = n_C + n_S = a + b = \dfrac{9,6}{32} = 0,3(2)\\ (1)(2)\Rightarrow a = 0,2 ; b = 0,1\\ \%m_C = \dfrac{0,2.12}{5,6}.100\% =42,86\%\\ \%m_S = 100\%-42,86\% = 57,14\%\)
\(n_{CO_2} = n_C = 0,2(mol)\\ n_{SO_2} = n_S = 0,1(mol)\\ \%V_{CO_2} = \dfrac{0,2}{0,2 + 0,1}.100\% = 66,67\%\\ \%V_{SO_2} = 100\%-66,67\% = 33,33\%\\ m_{hh\ sau\ pư} = m_C + m_S + m_{O_2} = 5,6 + 9,6 = 15,2(gam)\\ \%m_{CO_2} = \dfrac{0,2.44}{15,2}.100\% = 57,89\%\\ \%m_{SO_2} = 100\% -57,89\% = 42,11\%\)
Có \(\left\{{}\begin{matrix}n_{H_2}+n_{C_2H_2}=\dfrac{17,92}{22,4}=0,8\\\dfrac{2.n_{H_2}+26.n_{C_2H_2}}{n_{H_2}+n_{C_2H_2}}=0,5.28=14\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}n_{H_2}=0,4\left(mol\right)\\n_{C_2H_2}=0,4\left(mol\right)\end{matrix}\right.\)
\(n_{O_2}=\dfrac{51,2}{32}=1,6\left(mol\right)\)
PTHH: 2H2 + O2 --to--> 2H2O
0,4-->0,2
2C2H2 + 5O2 --to--> 4CO2 + 2H2O
0,4----->1------------>0,8
=> Y chứa \(\left\{{}\begin{matrix}CO_2:0,8\left(mol\right)\\O_{2\left(dư\right)}:0,4\left(mol\right)\end{matrix}\right.\)
=> \(\overline{M}_Y=\dfrac{0,8.44+0,4.32}{0,8+0,4}=40\left(g/mol\right)\)
\(\overline{M}_X=14\left(g/mol\right)\)
=> \(d_{X/Y}=\dfrac{14}{40}=0,35\)
Ta có: \(n_{Cu}=\dfrac{8,32}{64}=0,13\left(mol\right)\)
\(n_{O_2}=\dfrac{6,81725}{24,79}=0,275\left(mol\right)\)
PT: \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
\(2Cu+O_2\underrightarrow{t^o}2CuO\)
Theo PT: \(n_{O_2}=\dfrac{2}{3}n_{Fe}+\dfrac{1}{2}n_{Cu}\) \(\Rightarrow n_{Fe}=0,315\left(mol\right)\)
\(\Rightarrow a=0,315.56=17,64\left(g\right)\)
Theo PT: \(\left\{{}\begin{matrix}n_{Fe_3O_4}=\dfrac{1}{3}n_{Fe}=0,105\left(mol\right)\\n_{CuO}=n_{Cu}=0,13\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%n_{Fe_3O_4}=\dfrac{0,105}{0,105+0,13}.100\%\approx44,7\%\\\%n_{CuO}\approx55,3\%\end{matrix}\right.\)
\(\overline{M_{hh}}=\dfrac{m_{Fe_3O_4}+n_{CuO}}{n_{Fe_3O_4}+n_{CuO}}=\dfrac{0,105.232+0,13.80}{0,105+0,13}=147,91\left(g/mol\right)\)