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a, \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
\(Cu+2H_2SO_{4\left(đ\right)}\underrightarrow{t^o}CuSO_4+SO_2+2H_2O\)
b, Ta có: \(n_{H_2}=\dfrac{2,479}{24,79}=0,1\left(mol\right)\)
\(n_{Zn}=n_{H_2}=0,1\left(mol\right)\)
\(n_{SO_2}=\dfrac{2,9748}{24,79}=0,12\left(mol\right)\)
\(n_{Cu}=n_{SO_2}=0,12\left(mol\right)\)
\(\Rightarrow m=m_{Zn}+m_{Cu}=0,1.65+0,12.64=14,18\left(g\right)\)
Có: \(n_{H_2SO_{4\left(đ\right)}}=2n_{SO_2}=0,24\left(mol\right)\Rightarrow x=m_{ddH_2SO_4\left(đ\right)}=\dfrac{0,24.98}{98\%}=24\left(g\right)\)
a, \(Cu+2H_2SO_{4\left(đ\right)}\underrightarrow{t^o}CuSO_4+SO_2+2H_2O\)
b, \(n_{SO_2}=\dfrac{3,7185}{24,79}=0,15\left(mol\right)\)
Theo PT: \(n_{Cu}=n_{CuSO_4}=n_{SO_2}=0,15\left(mol\right)\)
\(\Rightarrow m_{Cu}=0,15.64=9,6\left(g\right)=m\)
Theo PT: \(n_{H_2SO_4}=2n_{SO_2}=0,3\left(mol\right)\Rightarrow C\%_{H_2SO_4}=\dfrac{0,3.98}{200}.100\%=14,7\%=x\)
Ta có: m dd sau pư = 9,6 + 200 - 0,15.64 = 200 (g)
\(\Rightarrow C\%_{CuSO_4}=\dfrac{0,15.160}{200}.100\%=12\%\)
a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(ZnO+2HCl\rightarrow ZnCl_2+H_2O\)
b, Ta có: 65nZn + 81nZnO = 17,85 (1)
Theo PT: \(n_{ZnCl_2}=n_{Zn}+n_{ZnO}=\dfrac{34}{136}=0,25\left(mol\right)\) (2)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{Zn}=0,15\left(mol\right)\\n_{ZnO}=0,1\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow m_{ZnO}=0,1.81=8,1\left(g\right)\)
c, \(n_{HCl}=2n_{ZnCl_2}=0,5\left(mol\right)\) \(\Rightarrow V_{HCl}=\dfrac{0,5}{1,5}=\dfrac{1}{3}\left(l\right)=\dfrac{1000}{3}\left(ml\right)\)
\(n_{H_2}=n_{Zn}=0,15\left(mol\right)\Rightarrow V_{H_2}=0,15.24,79=3,7185\left(l\right)\)
Ta có: \(n_{H^+}=n_{HCl}=0,04.0,75=0,03\left(mol\right)\)
\(n_{OH^-}=2n_{Ba\left(OH\right)_2}+n_{KOH}=2.0,16.0,08+0,16.0,04=0,032\left(mol\right)\)
PT: \(H^++OH^-\rightarrow H_2O\)
____0,03______0,03 (mol)
\(\Rightarrow n_{OH^-\left(dư\right)}=0,032-0,03=0,002\left(mol\right)\)
\(\Rightarrow\left[OH^-\right]=\dfrac{0,002}{0,04+0,16}=0,01\left(M\right)\)
\(\Rightarrow\left[H^+\right]=\dfrac{10^{-14}}{0,01}=10^{-12}\left(M\right)\)
\(\Rightarrow pH=-log\left[H^+\right]=12\)
a, \(H_2+CuO\underrightarrow{t^o}Cu+H_2O\)
b, \(n_{Cu}=\dfrac{0,32}{64}=0,005\left(mol\right)\)
\(n_{CuO}=n_{Cu}=0,005\left(mol\right)\Rightarrow m_{CuO}=0,005.80=0,4\left(g\right)\)
c, \(n_{H_2}=n_{Cu}=0,005\left(mol\right)\Rightarrow V_{H_2}=0,005.24,79=0,12395\left(l\right)\)
d, \(n_{H_2O}=n_{Cu}=0,005\left(mol\right)\Rightarrow m_{H_2O}=0,005.18=0,09\left(g\right)\)
\(n_{Fe_2O_3}=\dfrac{11,2}{160}=0,07mol\\ Fe_2O_3+3CO\xrightarrow[t^0]{}2Fe+3CO_2\\ n_{Fe}=2.0,07=0,14mol\\ m_{Fe}=0,14.56=7,84g\)
\(n_{Fe_2O_3}=\dfrac{11,2}{160}=0,07\left(mol\right)\)
PTHH :
\(Fe_2O_3+3CO\rightarrow2Fe+3CO_2\)
0,07 0,14
\(m_{Fe}=0,14.56=7,84\left(g\right)\)
CuSO4 → KL pư hết.
Gọi: \(\left\{{}\begin{matrix}n_{Zn}=a\left(mol\right)\\n_{Fe}=b\left(mol\right)\end{matrix}\right.\) ⇒ m = 65a + 56b (1)
BT e, có: 2a + 2b = 2nCu ⇒ nCu = a + b (mol) ⇒ m = 64(a + b) (2)
Từ (1) và (2) ⇒ 65a + 56b = 64a + 64b
⇒ a = 8b
\(\Rightarrow\%m_{Zn}=\dfrac{65a}{65a+56b}.100\%=\dfrac{65.8b}{65.8b+56b}.100\%\approx90,28\%\)
Please