PTHH: \(2Al+3CuSO_4\rightarrow Al_2\left(SO_4\right)_3+3Cu\)
a) Gọi số mol Nhôm p/ứ là 2x (mol) \(\Rightarrow n_{Cu}=3x\left(mol\right)\)
Ta có: \(10-27\cdot2x+64\cdot3x=11,38\) \(\Rightarrow x=0,01\left(mol\right)\)
\(\Rightarrow m_{Cu}=3\cdot0,01\cdot64=1,92\left(g\right)\)
b) Theo PTHH: \(\left\{{}\begin{matrix}n_{CuSO_4\left(dư\right)}=0,2-0,03=0,17\left(mol\right)\\n_{Al_2\left(SO_4\right)_3}=0,01\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}C_{M_{CuSO_4\left(dư\right)}}=\dfrac{0,17}{0,5}=0,34\left(M\right)\\C_{M_{Al_2\left(SO_4\right)_3}}=\dfrac{0,01}{0,5}=0,02\left(M\right)\end{matrix}\right.\)
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