nH2 = \(\frac{8,96}{22,4}\) =0,4(mol)

Goi x,y,z lan luot la so mol cua Mg, Al, Cu

PTHH: Mg + 2HCl → MgCl2 + H2↑

x → x (mol)

2Al + 6HCl → 2AlCl3 + 3H2↑

y → 1,5y(mol)

- DD A: MgCl2, AlCl3 , HCl du

- Chat ran B: Cu

2Cu + O2 → 2CuO

z → z(mol)

Ta co he PT: \(\left\{{}\begin{matrix}24x+27y+64z=10\\x+1,5y=0,4\\80z=2,75\end{matrix}\right.\)

( an may tinh mode 52 de giai he PT)

Giai he PT: x = 0,1; y = 0,2 ; z \(\approx\) 0,03

→ mMg = 24x = 24.0,1 = 2,4 (g)

mAl = 27y = 27.0,2 = 5,4 (g)

mCu = 64z = 64.0,03 = 1,92 (g)

13 - (x-5) = 3 + 2x

13 - x + 5 = 3 + 2x

13 -x  = 3-5 + 2x

13 - x = -2 +2x

13 - x + 2 = 2x

13 + 2 = 2x + x

15       = 3x

x         = 15 / 3

x         = 5

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