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\(n_{H_2}=\dfrac{2,479}{24,79}=0,1\left(mol\right)\)
PTHH:
2K + 2H2O ---> 2KOH + H2
0,2<---------------0,2<----0,1
=> \(\left\{{}\begin{matrix}m_K=0,2.39=7,8\left(g\right)\\m_{K_2O}=12,5-7,8=4,7\left(g\right)\end{matrix}\right.\)
\(n_{K_2O}=\dfrac{4,7}{94}=0,05\left(mol\right)\)
PTHH: K2O + H2O ---> 2KOH
0,05-------------->0,1
=> mKOH = (0,2 + 0,1).56 = 16,8 (g)
Đặt \(\left\{{}\begin{matrix}n_{Fe_2O_3}=x\left(mol\right)\\n_{PbO}=y\left(mol\right)\end{matrix}\right.\)
\(PTHH:Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\\ \left(mol\right)......x\rightarrow...3x......2x.....3x\\ PTHH:PbO+H_2\underrightarrow{t^o}Pb+H_2O\\ \left(mol\right)......y\rightarrow.y.....y......y\\ m_{Fe_2O_3}+m_{PbO}=\Sigma m_{hh}\\ \Leftrightarrow160x+223y=76,6\left(1\right)\\ m_{Fe}+m_{Pb}=\Sigma m_{kl}\\ \Leftrightarrow56.2x+207y=63,8\\ \Leftrightarrow112x+207y=63,8\left(2\right)\\ \xrightarrow[\left(1\right)]{\left(2\right)}\left\{{}\begin{matrix}160x+223y=76,6\\112x+207y=63,8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,2\\y=0,2\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\%m_{Fe_2O_3}=\dfrac{0,2.160}{76,6}.100\%=41,8\%\\\%m_{PbO}=100\%-41,8\%=58,2\%\end{matrix}\right.\)
\(\Sigma n_{H_2}=3x+y=3.0,2+0,2=0,8\left(mol\right)\\ \Sigma V_{H_2}=0,8.22,4=17,92\left(l\right)\)
Câu c là H2 chứ bạn
a, \(2Na+2H_2O\rightarrow2NaOH+H_2\)
\(Na_2O+H_2O\rightarrow2NaOH\)
Ta có: \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
Theo PT: \(n_{Na}=2n_{H_2}=0,3\left(mol\right)\Rightarrow m_{Na}=0,3.23=6,9\left(g\right)\)
\(\Rightarrow m_{Na_2O}=13,1-6,9=6,2\left(g\right)\)
b, \(n_{Na_2O}=\dfrac{6,2}{62}=0,1\left(mol\right)\)
Theo PT: \(n_{NaOH}=n_{Na}+2n_{Na_2O}=0,5\left(mol\right)\)
Ta có: m dd sau pư = 13,1 + 200 - 0,15.2 = 212,8 (g)
\(\Rightarrow C\%_{NaOH}=\dfrac{0,5.40}{212,8}.100\%\approx9,4\%\)
\(2Na+2H_2O\rightarrow2NaOH+H_2\\ Na_2O+H_2O\rightarrow2NaOH\\ n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ n_{Na}=2.0,3=0,6\left(mol\right)\\ a,m_{Na}=0,6.23=13,8\left(g\right)\\ m_{Na_2O}=26,2-13,8=12,4\left(g\right)\\b, n_{Na_2O}=\dfrac{12,4}{62}=0,2\left(mol\right)\\ n_{NaOH\left(tổng\right)}=n_{Na}+2.n_{Na_2O}=0,6+\dfrac{12,4}{62}=0,8\left(mol\right)\\ m_{c.tan}=m_{NaOH}=0,8.40=32\left(g\right)\\ c,m_{ddNaOH}=m_{hh}+m_{H_2O}-m_{H_2}=26,2+200-0,3.2=225,6\left(g\right)\\ C\%_{ddNaOH}=\dfrac{32}{225,6}.100\approx14,185\%\)
`a)Fe + 2HCl -> FeCl_2 + H_2 \uparrow`
`0,1` `0,1` `(mol)`
`Cu + HCl -xx->`
`b)n_[H_2]=[2,479]/[22,4]=0,1 (mol)`
`m_[Fe]=0,1.56=5,6(g)`
`=>m_[Cu]=10-5,6=4,4(g)`
`c)%m_[Fe]=[5,6]/10 .100=56%`
`%m_[Cu]=100-56=44%`
`d)` Dung dịch sau phản ứng có làm đổi màu quỳ tím. Vì: `HCl` dư nên sau phản ứng quỳ tím đổi màu đỏ.
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!
\(n_{H_2}=\dfrac{3,36}{22,4}=0,15mol\)
\(\left\{{}\begin{matrix}n_{Al}=x\left(mol\right)\\n_{Ag}=y\left(mol\right)\end{matrix}\right.\Rightarrow27x+108y=4,2\left(1\right)\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
0,1 0,05 0,15
\(\Rightarrow m_{Al}=0,1\cdot27=2,7g\)
\(\Rightarrow m_{Ag}=4,2-2,7=1,5g\)
a)\(\%m_{Al}=\dfrac{2,7}{4,2}\cdot100\%=64,28\%\)
\(\%m_{Ag}=100\%-64,28\%=35,72\%\)
b)\(m_{muối}=0,05\cdot342=17,1g\)
Đổi: 250 cm3 = 0,25 dm3 = 0,25 lít
nH2 = \(\dfrac{0,25}{22,4}\) mol
Pt: 2K + 2H2O --> 2KOH + H2
\(\dfrac{5}{224}\) mol<--------------------\(\dfrac{0,25}{22,4}\) mol
mK = \(\dfrac{5}{224}.39=0,87\left(g\right)\)
mK2O = mhh - mK = 18 - 0,87 = 17,13 (g)