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a, PT: \(C+O_2\underrightarrow{t^o}CO_2\)
\(S+O_2\underrightarrow{t^o}SO_2\)
b, Giả sử: \(\left\{{}\begin{matrix}n_C=x\left(mol\right)\\n_S=y\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow12x+32y=5,6\left(1\right)\)
Ta có: \(n_{O_2}=\dfrac{9,6}{32}=0,3\left(mol\right)\)
Theo PT: \(\Sigma n_{O_2}=n_C+n_S=x+y\left(mol\right)\)
\(\Rightarrow x+y=0,3\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,2\left(mol\right)\\y=0,1\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_C=0,2.12=2,4\left(g\right)\\m_S=0,1.32=3,2\left(g\right)\end{matrix}\right.\)
c, Ta có: \(\left\{{}\begin{matrix}\%m_C=\dfrac{2,4}{5,6}.100\%\approx42,9\%\\\%m_S\approx57,1\%\end{matrix}\right.\)
d, Phần này đề yêu cầu tính theo khối lượng mol hả bạn?
a, \(2KClO_3\underrightarrow{t^0}2KCl+3O_2\)
\(2KMnO_4\underrightarrow{t^0}K_2MnO_4+MnO_2+O_2\)gdf
b, Gọi số mol KClO3 và KMnO4 lần lượt là x,y ( mol ) ( x,y > 0 )
\(2KClO_3\underrightarrow{t^0}2KCl+3O_2\)
x(mol)......................\(\dfrac{3}{2}x\left(mol\right)\)
\(2KMnO_4\underrightarrow{t^0}K_2MnO_4+MnO_2+O_2\)
y(mol)..............................................\(\dfrac{1}{2}y\left(mol\right)\)
\(n_{O_2}=\dfrac{V}{22,4}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)
Tổng số mol O2 : \(\dfrac{3}{2}x+\dfrac{1}{2}y=0,05\left(1\right)\)
\(m_{KClO_3}=n.M=122,5x\left(g\right)\)
\(m_{KMnO_4}=n.M=158y\left(g\right)\)
\(\Rightarrow122,5x+158y=8,77\left(2\right)\)
Từ (1)(2) ,có :\(\left\{{}\begin{matrix}\dfrac{3}{2}x+\dfrac{1}{2}y=0,05\\122,5x+158y=8,77\end{matrix}\right.\) ( bấm máy tính là ra )
\(\Rightarrow\left\{{}\begin{matrix}x=0,02\\y=0,04\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{KClO_3}=122,5.x=122,5.0,02=2,45\left(g\right)\\m_{KMnO_4}=158y=158.0,04=6,32\left(g\right)\end{matrix}\right.\)
\(\%m_{KClO_3}=\dfrac{2,45}{8,77}.100\%=27\%\)
\(\%m_{KMnO_4}=\dfrac{6,32}{8,77}.100\%=73\%\)
nO2=1,12/22,4=0,05(mol)
2KClO3--->2KCl+3O2
x_______________3/2x
2KMnO4--->K2MnO4+MnO2+O2
y__________________________1/2y
Hệ pt:
\(\left\{{}\begin{matrix}122,5x+158y=8,77\\1,5x+0,5y=0,05\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,02\\y=0,04\end{matrix}\right.\)
=>mKClO3=0,02.122,5=2,45(g)
=>%mKClO3=2,45/8,77.100%~27,9%
=>%mKMnO4=100%-27,9%=72,1%
\(n_{H_2}=\dfrac{5,6}{22,4}=0,25(mol)\\ a,PTHH:Fe+2HCl\to FeCl_2+H_2\\ 2Al+6HCl\to 2AlCl_3+3H_2\)
\(b,\) Đặt \(n_{Fe}=x(mol);n_{Al}=y(mol)\)
\(\Rightarrow 56x+27y=8,3(1)\)
Theo PTHH: \(x+1,5y=0,25(2)\)
\((1)(2)\Rightarrow x=y=0,1(mol)\\ \Rightarrow \%_{Fe}=\dfrac{0,1.56}{8,3}.100\%=67,47\%\\ \%_{Al}=100\%-67,47\%=32,53\%\)
\(a,n_{H_2}=\dfrac{1,68}{22,4}=0,075\left(mol\right)\)
PTHH: \(Mg+H_2SO_4\rightarrow MgSO_4+H_2\uparrow\)
0,075<-------------------------0,075
Cu không phản ứng với H2SO4 loãng
b, \(m_{Mg}=0,075.24=1,8\left(g\right)\)
\(\rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{1,8}{8}.100\%=22,5\%\\\%m_{Cu}=100\%-22,5\%=77,5\%\end{matrix}\right.\)
\(n_{H_2}=\dfrac{1,68}{22,4}=0,075\left(mol\right)\\ pthh:Mg+H_2SO_4\rightarrow MgSO_4+H_2\)
0,075 0,075
\(Cu+H_2SO_4-x->\)
\(m_{Mg}=0,075.24=1,6\left(g\right)\\ m_{Cu}=8-1,6=6,4\left(g\right)\)
\(\%m_{Cu}=\dfrac{6,4}{8}.100\%=80\%\\
\%m_{Mg}=100-80\%=20\%\)
a)
Mg + 2HCl --> MgCl2 + H2
b)
\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
PTHH: Mg + 2HCl --> MgCl2 + H2
0,3<-----------------0,3
=> mMg = 0,3.24 = 7,2 (g)
=> mAg = 10,4 - 7,2 = 3,2 (g)
c) \(\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{7,2}{10,4}.100\%=69,23\%\\\%m_{Ag}=\dfrac{3,2}{10,4}.100\%=30,77\%\end{matrix}\right.\)
a, Cu không tác dụng với dd HCl.
PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
b, Ta có: \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
Theo PT: \(n_{Zn}=n_{H_2}=0,2\left(mol\right)\)
\(\Rightarrow m_{Zn}=0,2.65=13\left(g\right)\)
\(\Rightarrow m_{Cu}=19,4-13=6,4\left(g\right)\)
c, Ta có: \(\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{13}{19,4}.100\%\approx67,01\%\\\%m_{Cu}\approx32,99\%\end{matrix}\right.\)
Bạn tham khảo nhé!
Gọi x,y lần lượt là số mol của Fe2O3, CuO
nCO = \(\dfrac{2,016}{22,4}=0,09\) mol
Pt: Fe2O3 + 3CO --to--> 2Fe + 3CO2
........x............3x...............2x
.....CuO + CO --to--> Cu + CO2
.......y..........y.................y
Ta có hệ pt:\(\left\{{}\begin{matrix}160x+80y=5,6\\3x+y=0,09\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,02\\y=0,03\end{matrix}\right.\)
=> %
nCO=2,016/22,4=0,09(mol)
Fe2O3+3CO--->2Fe+3CO2
x_______3x_____2x
CuO+CO--->Cu+CO2
y_____y_____y
Hệ pt:
\(\left\{{}\begin{matrix}160x+80y=5,6\\3x+y=0,09\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,02\\y=0,03\end{matrix}\right.\)
=>mFe2O3=0,02.160=3,2(g)
=>%fe2O3=3,2/5,6.100~57%
=>%CuO=100-57=43%
mFe=0,04.56=2,24(g)
mCu=0,03.64=1,92(g)
=>%mFe=2,24/4,16.100~54%
=>%mCu=100-54=46%