Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1.
2M + 2xHCl \(\rightarrow\)2MClx + xH2
nH2=\(\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
Theo PTHH ta có:
nM=\(\dfrac{2}{x}\)nH2=\(\dfrac{0,6}{x}\)
MM=\(\dfrac{5,4}{\dfrac{0,6}{x}}=9x\)
Với x=3 thì MM=27
Vậy M là Al
Mg + 2 H2SO4 (đ) -to-> MgSO4 + SO2 + 2 H2O
x_________2x__________________x(mol)
2 Fe + 6 H2SO4(đ) -to-> Fe2(SO4)3 + 3 SO2 + 6 H2O
y______3y_____________________1,5y(mol)
Ta có hpt:
\(\left\{{}\begin{matrix}24x+56y=18,4\\x+1,5y=0,6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,3\\y=0,2\end{matrix}\right.\)
=> mMg= 0,3.24=7,2(g)
=> %mMg= (7,2/18,4).100=39,13%
=>%mFe= 60,87%
b) nH2SO4(tổng)=2x+3y=2.0,3+3.0,2=1,2(mol)
VddH2SO4=1,2/2=0,6(l)
a) Đặt \(\left\{{}\begin{matrix}n_{Fe}=a\left(mol\right)\\n_{Mg}=b\left(mol\right)\end{matrix}\right.\) \(\Rightarrow56a+24b=18,4\) (1)
Ta có: \(n_{SO_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)
Bảo toàn electron: \(3a+2b=0,6\cdot2\) (2)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}a=0,2\\b=0,3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,2\cdot56}{18,4}\cdot100\%\approx60,87\%\\\%m_{Mg}=39,13\%\end{matrix}\right.\)
b) Ta có: \(\left\{{}\begin{matrix}n_{Fe_2\left(SO_4\right)_3}=\dfrac{1}{2}n_{Fe}=0,1\left(mol\right)\\n_{MgSO_4}=n_{Mg}=0,3\left(mol\right)\end{matrix}\right.\)
Bảo toàn nguyên tố: \(n_{H_2SO_4}=n_{SO_2}+3n_{Fe_2\left(SO_4\right)_3}+n_{MgSO_4}=1,2\left(mol\right)\)
\(\Rightarrow V_{H_2SO_4}=\dfrac{1,2}{2}=0,6\left(l\right)=600\left(ml\right)\)
2Al + 3H2SO4 \(\rightarrow\)Al2(SO4)3 + 3H2 (1)
Zn + H2SO4 \(\rightarrow\)ZnSO4 + H2 (2)
a;nH2=\(\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
Đặt nAl=a
nZn=b
Ta có:
\(\left\{{}\begin{matrix}27a+65b=11,9\\\dfrac{3}{2}a+b=0,4\end{matrix}\right.\)
=>a=0,2;b=0,1
mAl=27.0,2=5,4(g)
%mAl=\(\dfrac{5,4}{11,9}.100\%=45,4\%\)
%mZn=100-45,4=54,6%
b;Theo PTHH 1 và 2 ta có:
nH2=nH2SO4=0,4(mol)
VH2SO4=\(\dfrac{0,4}{0,5}=0,8\left(lít\right)\)
\(n_{H_2}=0,4\left(mol\right)\)
\(Zn+H_2SO_4-->ZnSO_4+H_2\uparrow\)
x.........x...............................x............x
\(2Al+3H_2SO_4-->Al_2\left(SO_4\right)_3+3H_2\uparrow\)
y.........1,5y.......................0,5y.................1,5y
\(\left\{{}\begin{matrix}65x+27y=11,9\\x+1,5y=0,4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)
\(\%Zn=\dfrac{0,1.65}{11,9}.100\%\approx54,62\%\)
\(\%Al=100\%-54,62\%=45,38\%\)
b) nH2=nH2SO4=0,4(mol)
\(V_{ddH_2SO_4}=\dfrac{0,4}{0,5}=0,8\left(l\right)\)
a. \(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
0,25 ..... 0,5 ................... 0,25 (mol)
\(m_{Mg}=0,25.24=6\left(g\right)\)
\(\rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{6}{10}.100\%=60\%\\\%m_{MgO}=100\%-60\%=40\%\end{matrix}\right.\)
\(MgO+2HCl\rightarrow MgCl_2+H_2O\)
0,1 ....... 0,2 (mol)
\(n_{HCl}=0,25+0,1=0,35\left(mol\right)\)
\(C_M\left(HCl\right)=\dfrac{0,35}{0,1}=3,5\left(M\right)\)
Ta có: mO (trong oxit) = 10 - 8,4 = 1,6 (g)
\(n_{Fe}=\dfrac{8,4}{56}=0,15\left(mol\right)\)
\(n_O=\dfrac{1,6}{16}=0,1\left(mol\right)\)
Giả sử: \(n_{SO_2}=x\left(mol\right)\)
Theo ĐLBT mol e, có: 0,15.3 = 0,1.2 + 2x
⇒ x = 0,125 (mol)
\(\Rightarrow V_{SO_2}=0,125.22,4=2,8\left(l\right)\)
Bạn tham khảo nhé!
\(n_{Fe}=\dfrac{8.4}{56}=0.15\left(mol\right)\)
\(m_{O_2}=10-8.4=1.6\left(g\right)\)
\(n_{O_2}=0.1\left(mol\right)\)
Bảo toàn e :
\(n_{SO_2}=\dfrac{3\cdot0.15+0.1\cdot4}{2}=0.425\left(mol\right)\)
\(V_{SO_2}=0.425\cdot22.4=9.52\left(l\right)\)