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\(a.n_{H_2}=\dfrac{1,2.10^{23}}{6.10^{23}}=0,2\left(mol\right)\\ n_{SO_2}=\dfrac{6,4}{64}=0,1\left(mol\right)\\ V_X=\left(1,5+2,5+0,2+0,1\right).22,4=96,32\left(l\right)\\b. m_X=1,5.32+2,5.28+0,2.2+6,4=124,8\left(g\right)\)
a.nH2=1,2.10236.1023=0,2(mol)nSO2=6,464=0,1(mol)VX=(1,5+2,5+0,2+0,1).22,4=96,32(l)b.mX=1,5.32+2,5.28+0,2.2+6,4=124,8(g)
Bài 1 :
a,
- 5,6g Fe.
\(\Rightarrow n_{Fe}=\frac{5,6}{56}=0,1\left(mol\right)\)
- 4,05g Al.
\(n_{Al}=\frac{4,05}{27}=0,15\left(mol\right)\)
- 7,8g Zn
\(n_{Zn}=\frac{65}{7,8}=0,12\left(mol\right)\)
b,
b. Tính thể khí (đktc) của:
- 0,5 mol CO2.
\(V_{CO2}=0,5.22,4=11,2\left(l\right)\)
- 0,75 mol N2.
\(V_{N2}=0,75.22,4=16,8\left(l\right)\)
- 0,3 mol CO.
\(V_{CO}=0,3.22,4=6,72\left(l\right)\)
c,
- 0,5 mol CO2; 0,75 mol N2; 0,3 mol CO.
\(V_{CO2}=0,5.22,4=11,2\left(l\right),m_{CO2}=0,5.44=22\left(g\right)\)
\(V_{N2}=0,75.22,4=16,8\left(l\right);m_{N2}=0,75.14=21\left(g\right)\)
\(V_{CO}=0,3.22,4=6,72\left(l\right),m_{CO}=0,3.8,4\left(g\right)\)
- 0,25 mol CO2; 0,5 mol N2; 0,35 mol CO.
\(V_{CO2}=0,25.22,4=5,6\left(l\right);m_{CO2}=0,24.44=10,56\left(g\right)\)
\(V_{N2}=0,5.22,4=11,2\left(l\right);m_{N2}=28.0,5=9\left(g\right)\)
\(V_{CO}=0,35.22,4=7,84\left(l\right);m_{CO}=0,35.28=9,8\left(g\right)\)
- 0,05 mol CO2; 0,7 mol N2; 0,6 mol CO. (Tương tự nha )
2Al + 3H2SO4 →Al2(SO4)3 + 3H2 (1)
Zn + H2SO4 →ZnSO4 + H2 (2)
a;nH2=\(\dfrac{8,96}{22,4}\)=0,4(mol)
Đặt nAl=a
nZn=b
Ta có:
\(\left\{{}\begin{matrix}27x+65y=11,9\\1,5x+y=0,4\end{matrix}\right.\)
=>a=0,2;b=0,1
mAl=27.0,2=5,4(g)
%mAl=\(\dfrac{5,4}{11,9}.100\)=54,4%
%mZn=54,6%
Ta có kim loại + H2SO4 → muối + H2
nH2 = 0,4 mol
Bảo toàn nguyên tố H có nH2 = nH2SO4 = 0,4 mol
Bảo toàn khối lượng có mkim loại + mH2SO4 = mH2 + mmuối → 11,9 + 0,4.98 = 0,4.2 + m → m = 50,3
Gọi \(n_{Fe}=a\left(mol\right)\rightarrow n_{Mg}=\dfrac{1}{1}.a=a\left(mol\right)\)
\(\rightarrow n_{Zn}=0,3-a-a=0,3-2a\left(mol\right)\)
\(\rightarrow65\left(0,3-2a\right)+56a+24a=13\\ \Leftrightarrow a=0,13\\ \Rightarrow\left\{{}\begin{matrix}n_{Fe}=n_{Mg}=0,13\left(mol\right)\\n_{Zn}=0,3-0,13.2=0,04\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{56.0,13}{13}.100\%=56\%\\\%m_{Mg}=\dfrac{24.0,13}{13}.100\%=24\%\\\%m_{Zn}=100\%-56\%-25\%=20\%\end{matrix}\right.\)
PTHH:
Fe + 2HCl ---> FeCl2 + H2
Zn + 2HCl ---> ZnCl2 + H2
Mg + 2HCl ---> MgCl2 + H2
Theo pthh: nH2 = nkim loại = 0,3 (mol)
\(n_{CuO}=\dfrac{80}{80}=1\left(mol\right)\)
PTHH: CuO + H2 --to--> Cu + H2O
LTL: 1 > 0,3 => CuO dư
Chất rắn sau pư gồm: CuO dư, Cu
Theo pthh: nCuO (pư) = nCu = nH2 = 0,3 (mol)
=> mchất rắn = 80,(1 - 0,3) + 64.0,3 = 75,2 (g)
a) Gọi số mol Al, Zn là 2a, a (mol)
PTHH: 4Al + 3O2 --to--> 2Al2O3
2a-->1,5a---------->a
2Zn + O2 --to--> 2ZnO
a---->0,5a------->a
=> \(102a+81a=18,3\)
=> a = 0,1 (mol)
=> \(\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,2.27}{0,2.27+0,1.65}.100\%=45,378\%\\\%m_{Zn}=\dfrac{0,1.65}{0,2.27+0,1.65}.100\%=54,622\%\end{matrix}\right.\)
b) \(n_{O_2}=1,5a+0,5a=0,2\left(mol\right)\)
=> \(V_{O_2}=0,2.22,4=4,48\left(l\right)\)
a, \(\overline{M}=\dfrac{0,1.44+0,2.28}{0,1+0,2}\approx33,33\left(g/mol\right)\)
b, \(\overline{M}=\dfrac{0,2.28+0,3.2}{0,2+0,3}=12,4\left(g/mol\right)\)
c, \(\overline{M}=\dfrac{0,1.28+0,2.30+0,2.44}{0,1+0,2+0,2}=35,2\left(g/mol\right)\)
d, \(\overline{M}=\dfrac{0,2.56+0,1.24+0,1.27}{0,2+0,1+0,1}=40,75\left(g/mol\right)\)
1) mchất rắn = \(0,3\cdot56+0,7\cdot65=62,3\left(g\right)\)
2) Ta có: \(\left\{{}\begin{matrix}n_{O_2}=\frac{12,8}{32}=0,4\left(mol\right)\\n_{N_2}=\frac{44,8}{28}=1,6\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow V_{hh}=\left(0,4+1,6\right)\cdot22,4=44,8\left(l\right)\)