Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Câu 1:
a, PT: \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
\(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
Giả sử: \(\left\{{}\begin{matrix}n_{CuO}=x\left(mol\right)\\n_{Fe_2O_3}=y\left(mol\right)\end{matrix}\right.\)
⇒ 80x + 160y = 40 (1)
Ta có: \(n_{H_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)
Theo PT: \(\Sigma n_{H_2}=n_{CuO}+3n_{Fe_2O_3}=x+3y\left(mol\right)\)
⇒ x + 3y = 0,6 (2)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,3\left(mol\right)\\y=0,1\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{CuO}=0,3.80=24\left(g\right)\\m_{Fe_2O_3}=0,1.160=16\left(g\right)\end{matrix}\right.\)
b, Ta có: \(\left\{{}\begin{matrix}\%m_{CuO}=\dfrac{24}{40}.100\%=60\%\\\%m_{Fe_2O_3}=40\%\end{matrix}\right.\)
Câu 4:
PT: \(2Na+2H_2O\rightarrow2NaOH+H_2\)
Ta có: \(n_{Na}=\dfrac{6,9}{23}=0,3\left(mol\right)\)
\(n_{H_2O}=\dfrac{150}{18}=\dfrac{25}{3}\left(mol\right)\)
Xét tỉ lệ ta được H2O dư.
Theo PT: \(n_{NaOH}=n_{Na}=0,3\left(mol\right)\) \(\Rightarrow m_{NaOH}=0,3.40=12\left(g\right)\)
\(n_{H_2}=\dfrac{1}{2}n_{Na}=0,15\left(mol\right)\)
Ta có: m dd sau pư = mNa + mH2O - mH2 = 6,9 + 150 - 0,15.2 = 156,6 (g)
\(\Rightarrow C\%_{NaOH}=\dfrac{12}{156,6}.100\%\approx7,66\%\)
Bạn tham khảo nhé!
PTHH: 2Na +2H2O →2NaOH+ H2 ↑
\(+n_{Na}=\dfrac{6,9}{23}=0,3\left(mol\right)\)
Theo PTHH ta có:
\(+n_{NaOH}=n_{Na}=0,3\left(mol\right)\)
\(+n_{H_2}=\dfrac{1}{2}n_{Na}=0,15\left(mol\right)\)
\(+m_{NaOH}=0,3.40=12\left(gam\right)\)
\(+C\%_{NaOH}=\dfrac{12}{150+6,9-0,15.2}.100\%\approx7,66\%\)
\(n_{H_2SO_4}=0,25.2=0,5\left(mol\right)\)
Gọi \(\left\{{}\begin{matrix}n_{Al_2O_3}=x\left(mol\right)\\n_{CuO}=y\left(mol\right)\end{matrix}\right.\)
\(Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)
x----------> 3x --------> x
\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)
y --------> y --------> y
Có hệ phương trình
\(\left\{{}\begin{matrix}102x+80y=26,2\\3x+y=0,5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)
\(\%_{m_{Al_2O_3}}=\dfrac{102.0,1.100}{26,2}=38,93\%\)
\(\%_{m_{CuO}}=\dfrac{80.0,2.100}{26,2}=61,07\%\)
\(CM_{Al_2\left(SO_4\right)_3}=\dfrac{x}{0,25}=\dfrac{0,1}{0,25}=0,4M\)
\(CM_{CuSO_4}=\dfrac{y}{0,25}=\dfrac{0,2}{0,25}=0,8M\)
a, -Gọi số mol của CuO và Fe2O3 lần lượt là x, y ( mol )
PTKL : \(80x+160y=40\left(I\right)\)
\(CuO+H_2\rightarrow Cu+H_2O\)
..x.........x............
\(Fe_2O_3+3H_2\rightarrow2Fe+3H_2O\)
...y............3y......
=> \(n_{H_2}=x+3y=\dfrac{V}{22,4}=0,6\left(mol\right)\left(II\right)\)
- Giair I và II ta được : x = 0,3 , y = 0,1 ( mol )
=> \(\left\{{}\begin{matrix}mCuO=n.M=24\left(g\right)\\mFe2O3=mhh-mCuO=16\left(g\right)\end{matrix}\right.\)
b, \(\%CuO=\dfrac{m}{mhh}.100\%=60\%\)
=> %Fe2O3 =100% - %CuO = 40% .
Vậy ...
\(CuO+2HCl \to CuCl_2+H_2O\\ Al_2O_3+6HCl \to 2AlCl_3+3H_2O\\ n_{CuO}=a(mol)\\ n_{Al_2O_3}=b(mol)\\ n_{HCl}=2a+6b=0,08(1)\\ m_{muối}=135a+267b=4,02(2)\\ (1)(2)\\ a=b=0,01(mol)\\ m_{dd}=4,02+0,01.(80+102)=5,84g\\ C\%_{CuO}=\frac{0,01.80}{5,84}.100\%=13,7\%\\ C\%_{Al_2O_3}=\frac{0,01.102}{5,84}.100\%=17,4\%\)
Đặt \(\hept{\begin{cases}x=n_{CuO}\\y=n_{Al_2O_3}\end{cases}}\)
PTHH: \(CuO+2HCl\rightarrow CuCl_2+H_2O\)
\(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\)
Theo phương trình \(\hept{\begin{cases}x=n_{CuCl_2}\\2y=n_{AlCl_3}\end{cases}}\)
\(\rightarrow135x+133,5.2y=4,02\left(1\right)\)
\(n_{HCl}=0,8.0,1=0,08mol\)
\(\rightarrow2x+6y=0,08\left(2\right)\)
Từ (1) và (2) => x = y = 0,01 mol
\(\rightarrow\%m_{CuO}=\frac{0,01.80}{0,01.80+0,01.102}.100\%=43,96\%\)
\(\rightarrow\%m_{Al_2O_3}=100-43,96\%=56,04\%\)
Đặt:\(\hept{\begin{cases}x=n_{CuO}\\y=n_{al_2O_3}\end{cases}}\)
PTHH:\(CuO+2HCL\rightarrow CuCl_2+H_2O\)
\(Al_2O_3+6HCL\rightarrow2Alcl_3+3H_2O\)
Theo phương trình: \(\hept{\begin{cases}x=n_{CuCl_2}\\2y=n_{AlCl_3}\end{cases}}\)
\(\rightarrow135x+133,5\cdot2y=4,02\left(1\right)\)
\(n_{HCL}=0,8\cdot0,1=0,08mol\)
\(\rightarrow2x+6y=0,08\left(2\right)\)
\(\Rightarrow x=y=0,01mol\)
\(\rightarrow\%m_{CuO}=\frac{0,01\cdot80}{0,01\cdot80+0,01\cdot102}\cdot100=43,96\%\)
\(\rightarrow\%m_{al_2O_3}=100\%-43,96\%=56,04\%\)
Đặt \(\left\{{}\begin{matrix}n_{CuO}=x\left(mol\right)\\n_{Fe_2O_3}=y\left(mol\right)\end{matrix}\right.\)
\(m_{CuO}+m_{Fe_2O_3}=40\\ \Rightarrow80x+160y=40\left(1\right)\)
\(PTHH:CuO+H_2\underrightarrow{t^o}Cu+H_2O\\ \left(mol\right)......x\rightarrow.x\\ PTHH:Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\\ \left(mol\right).....y\rightarrow....3y\\ V_{H_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\\ \Rightarrow x+3y=0,6\left(2\right)\)
\(\xrightarrow[\left(2\right)]{\left(1\right)}\left\{{}\begin{matrix}80x+160y=40\\x+3y=0,6\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=0,3\\y=0,1\end{matrix}\right.\)
a) \(\left\{{}\begin{matrix}m_{CuO}=80.0,3=24\left(g\right)\\m_{Fe_2O_3}=40-24=16\left(g\right)\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}\%m_{CuO}=\dfrac{24}{40}.100\%=60\%\\\%m_{Fe_2O_3}=100\%-60\%=40\%\end{matrix}\right.\)
gọi x và y lần lượt là số mol của Al2O3 và CuO ( x nguyên dương )
\(Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)
x ---- -- > 3x x 3x (mol)
\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)
y y y y (mol)
\(n_{H_2SO_4}=\dfrac{49}{98}=0,5\left(mol\right)\)
theo đề ta có hpt:
\(\left\{{}\begin{matrix}102x+80y=26,2\\3x+y=0,5\end{matrix}\right.\)
=> x = 0,1 ; y = 0,2
=> \(mAl_2O_3=102.0,2=20,4\left(g\right)\)
=> \(\%mAl_2O_3=\dfrac{20,4.100}{26,2}=77,86\%\)
=> \(\%mCuO=100\%-77,86\%=22,14\%\)