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Ta có: \(n_{NO}+n_{NO_2}+n_{N_2}=\dfrac{22,4}{22,4}=1\left(mol\right)\left(1\right)\)
Mà: mX = 35,8 (g)
\(\Rightarrow30n_{NO}+46n_{NO_2}+28n_{N_2}=35,8\left(2\right)\)
Có: \(n_{Al}=\dfrac{32,4}{27}=1,2\left(mol\right)\)
\(n_{Cu}=\dfrac{22,4}{64}=0,35\left(mol\right)\)
BT e, có: 3nNO + nNO2 + 10nN2 = 3nAl + 2nCu = 4,3 (3)
Từ (1), (2) và (3) \(\Rightarrow\left\{{}\begin{matrix}n_{NO}=0,3\left(mol\right)\\n_{NO_2}=0,4\left(mol\right)\\n_{N_2}=0,3\left(mol\right)\end{matrix}\right.\)
⇒ nHNO3 = 4nNO + 2nNO2 + 12nN2 = 5,6 (mol)
Đáp án C.
9x = 0,11; x= 11/900 => V = 5x.22,4 = 1,368 (l)
Đặt $n_{NO}=2a(mol);n_{NO_2}=a(mol)$
Bảo toàn e ta có: $6a+a=0,1.3+0,25.3\Rightarrow a=0,15(mol)$
Do đó $n_{A}=0,15.3=0,45(mol)\Rightarrow V_A=10,08(l)$
tham khảo trong:
https://moon.vn/hoi-dap/hoa-tan-hoan-toan-hon-hop-gom-01-mol-fe-va-025-mol-al-vao-dung-dich-hno3-du-thu-duoc-530914
a) \(\left\{{}\begin{matrix}160n_{Fe_2O_3}+80n_{CuO}=24\\n_{Fe_2O_3}=n_{CuO}\end{matrix}\right.\Rightarrow n_{Fe_2O_3}=n_{CuO}=0,1\)
\(\left\{{}\begin{matrix}\%m_{Fe_2O_3}=\dfrac{160.0,1}{24}.100\%=66,67\%\\\%m_{CuO}=\dfrac{80.0,1}{24}.100\%=33,33\%\end{matrix}\right.\)
PTHH: Fe2O3 + 3H2SO4 --> Fe2(SO4)3 + 3H2O
0,1------>0,3-------->0,1
CuO + H2SO4 --> CuSO4 + H2O
0,1-->0,1---------->0,1
nCuSO4 = 0,1 (mol)
nFe2(SO4)3 = 0,1 (mol)
=> m = 0,1.160 + 0,1.400 = 56(g)
b) \(m_{H_2SO_4\left(pthh\right)}=\left(0,3+0,1\right).98=39,2\left(g\right)\)
=> mH2SO4(thực tế) = \(\dfrac{39,2.125}{100}=49\left(g\right)\)
c) \(n_{H_2SO_4}=\dfrac{49}{98}=0,5\left(mol\right)\)
=> nBaSO4 = 0,5 (mol)
=> mBaSO4 = 0,5.233 = 116,5(g)
a)\(\left\{{}\begin{matrix}n_{Cu}=x\left(mol\right)\\n_{Al}=y\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}64x+27y=9,1\\BTe:2x+3y=0,5\cdot1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,1\end{matrix}\right.\)
\(\%m_{Cu}=\dfrac{0,1\cdot64}{9,1}\cdot100\%=70,33\%\)
\(\%m_{Al}=100-70,33\%=29,67\%\)
b)\(\left\{{}\begin{matrix}n_{NO_2}+n_{NO}=0,5\\\dfrac{n_{NO_2}}{n_{NO}}=\dfrac{2}{1}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}n_{NO_2}=\dfrac{1}{3}\\n_{NO}=\dfrac{1}{6}\end{matrix}\right.\)
Gọi \(\left\{{}\begin{matrix}n_{Cu}=a\left(mol\right)\\n_{Al}=b\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}64a+27b=9,1\\BTe:2x+3y=\dfrac{1}{3}\cdot1+\dfrac{1}{6}\cdot3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=\dfrac{4}{115}\\b=\dfrac{527}{2070}\end{matrix}\right.\)
\(\%m_{Cu}=\dfrac{\dfrac{4}{115}\cdot64}{9,1}\cdot100\%=24,46\%\)
\(\%m_{Al}=100\%-24,46\%=75,54\%\)
\(n_{HNO_3}=2n_{NO_2}+4n_{NO}=2\cdot\dfrac{1}{3}+4\cdot\dfrac{1}{6}=\dfrac{4}{3}mol\)
Dạ cảm ơn nhiều ạ