\(\begin{array}{l} mCu = 6g\\ = > mZn + mAl = 31,7 - 6 = 25,7g\\ 2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\ Zn + 2HCl \to ZnC{l_2} + {H_2}\\ hh:Al(a\,mol),Zn(b\,mol)\\ n{H_2} = \dfrac{{17,36}}{{22,4}} = 0,775\,mol\\ 27a + 65b = 25,7\\ 1,5a + b = 0,775\\ = > a = 0,35;b = 0,25\\ \% mCu = \dfrac{6}{{31,7}} \times 100\% = 18,93\% \\ \% mAl = \dfrac{{0,35 \times 27}}{{31,7}} \times 100\% = 29,81\% \\ \% mZn = 100 - 29,81 - 18,93 = 51,26\% \end{array}\)