Question

Answer 1

a)

2X + 2nHCl → 2XCl_n  +  nH₂

nH₂ = 3,36:22,4 = 0,15 mol => nX = \(\dfrac{0,3}{n}\) mol

<=> M_X = \(\dfrac{m_X}{n_X}\)= 12n (g/mol)

=> Với n = 2 , M_x = 24(g/mol) là giá trị thỏa mãn

<=> X là magie (Mg)

b)

2Mg  +  O₂  → 2MgO

2Cu  +  O₂  → 2CuO

nMg= 3,6:24 = 0,15 = nMgO

nCu = 3,2:64= 0,05 = nCuO

=> mOxit = mMgO + mCuO = 0,15.40 + 0,05.80 = 10 gam

Answer 2

Bài 2:

X  +  nH₂O  →  X(OH)_n  +  \(\dfrac{n}{2}\)H₂

nH₂ = 5,6:22,4 = 0,25 mol => nX = \(\dfrac{0,5}{n}\) mol <=> mX = \(\dfrac{0,5.M_X}{n}\) = a gam

4X + nO₂    →  2X₂O_n

nO₂ = 4,48:22,4 = 0,2 mol => nX = \(\dfrac{0,8}{n}\)mol

<=> b = \(\dfrac{0,8.M_X}{n}\) (gam)

=> \(\dfrac{a}{b}\)= \(\dfrac{0,5.M_X}{n}\):\(\dfrac{0,8.M_X}{n}\)= \(\dfrac{5}{8}\)