1.44 g of Titanium (Ti) reacted with an excess of O2 and produced x g of a non-stoichiometric compound Ti1.44O1. The value of x in g is :

Question

1.44 g of Titanium (Ti) reacted with an excess of O2 and produced x g of a non-stoichiometric compound Ti1.44O1. The value of x in g is : (A) 1.44 (B) 2 (C) 1.77 (D) None of these

Tardigrade Admin · Accepted Answer

Ti + O2 arrow Ti1.44 O (1.44 g) (x) g Apply POAC to Ti (. (1.44 textg/48) .) × 1 = . ( textx textg/(. 1.44 × 48 .) + 16) . 1.44 x = (4 8 × 1 text. 4 4 + 1 6/4 8) = 1 text. 4 4 + (1 6/4 8) = 1.44 + 0.33 = 1.77 g.

Q. 1.44 g of Titanium (Ti) reacted with an excess of O₂ and produced x g of a non-stoichiometric compound Ti_1.44O₁. The value of x in g is :

1.44

2

1.77

None of these

Ti + O₂ $\to$ Ti_1.44 O

(1.44 g) (x) g

Apply POAC to Ti

$(\frac{1.44 g}{48}) \times 1 = {\frac{x g}{( 1.44 \times 48 ) + 16}} 1.44$

x = $\frac{48 \times 1 . 44 + 16}{48}$

= $1.44 + \frac{16}{48}$

= 1.44 + 0.33

= 1.77 g.

Q. 1.44 g of Titanium (Ti) reacted with an excess of O2 and produced x g of a non-stoichiometric compound Ti1.44O1. The value of x in g is :

1.44

2

1.77

None of these

Ti + O2 → Ti1.44 O (1.44 g) (x) g Apply POAC to Ti (481.44g​)×1={(1.44×48)+16xg​}1.44 x = 4848×1.44+16​ = 1.44+4816​ = 1.44 + 0.33 = 1.77 g.

Q. 1.44 g of Titanium (Ti) reacted with an excess of O₂ and produced x g of a non-stoichiometric compound Ti_1.44O₁. The value of x in g is :

Ti + O₂ $\to$ Ti_1.44 O

(1.44 g) (x) g

Apply POAC to Ti

$(\frac{1.44 g}{48}) \times 1 = {\frac{x g}{( 1.44 \times 48 ) + 16}} 1.44$

x = $\frac{48 \times 1 . 44 + 16}{48}$

= $1.44 + \frac{16}{48}$

= 1.44 + 0.33

= 1.77 g.