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a,\(m_{HCl}=200.10\%=20\left(g\right)\Rightarrow n_{HCl}=\dfrac{20}{36,5}=0,548\left(mol\right)\)
b,\(n_{H_2SO_4}=0,2.0,2=0,04\left(mol\right)\)
PTHH: \(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)
\(Cu+2H_2SO_{4\left(đ\right)}\xrightarrow[]{t^o}CuSO_4+SO_2\uparrow+2H_2O\)
Khí A là SO2
Ta có: \(n_{SO_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)=n_{Cu}\)
\(\Rightarrow\%m_{Cu}=\dfrac{0,05\cdot64}{10}\cdot100\%=32\%\) \(\Rightarrow\%m_{CuO}=68\%\)
\(\left\{{}\begin{matrix}n_{Mg}=a\left(mol\right)\\n_{MgO}=b\left(mol\right)\end{matrix}\right.\left(a,b>0\right)\\ a.PTHH:Mg+2HCl\rightarrow MgCl_2+H_2\\ a.......................2a.........a..........a\left(mol\right)\\ PTHH:MgO+2HCl\rightarrow MgCl_2+H_2O\\ b.....................2b.............b..............b\left(mol\right)\\ \rightarrow\left\{{}\begin{matrix}95a+95b=47,5\\a=0,2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,2\\b=0,3\end{matrix}\right.\\ \rightarrow\left\{{}\begin{matrix}m_{Mg}=0,2.24=4,8\left(g\right)\\m_{MgO}=0,3.40=12\left(g\right)\end{matrix}\right.\\ \)
\(b.m_{HCl}=\left(2a+2b\right).36,5=36,5\left(g\right)\\ m_{ddHCl}=\dfrac{36,5.100}{14,6}=250\left(g\right)\\ m_{ddsau}=250+4,8+12-0,2.2=266,4\left(g\right)\\ C_{\%ddMgCl_2}=\dfrac{47,5}{266,4}.100\approx17,83\%\)
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