\(n_{CuO}=\dfrac{8}{80}=0,1\left(mol\right)\\ PTHH:CuO+2HCl\rightarrow CuCl_2+H_2\uparrow\\ \left(mol\right).....0,1\rightarrow...0,2.......0,1......0,1\)
a) \(m_{CuCl_2}=0,1.135=13,5\left(g\right)\)
b) \(100ml=0,1l\)
\(C_{M_{CuCl_2}}=\dfrac{0,1}{0,1}=1\left(M\right)\)
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