If 1.5 moles of oxygen combines with Al to form Al2O3, the mass of Al in g [Atomic mass of Al = 27] used in the reaction is

Question

If 1.5 moles of oxygen combines with Al to form Al2O3, the mass of Al in g [Atomic mass of Al = 27] used in the reaction is (A) 2.7 (B) 54 (C) 40.5 (D) 81

Tardigrade Admin · Accepted Answer

underset text 4moles 4Al + underset text 3moles 3 O 2 longrightarrow underset2 text moles 2 Al 2 O 3 3 moles of O 2 combine with 4 moles of Al ∴ 1.5 moles of O 2 will combine with x mol of Al x =(4/3) × 1.5=2 mol Thus mass of Al used in the reaction =2 × 27 g =54 g

Q. If $1.5$ moles of oxygen combines with $A l$ to form $A l_{2} O_{3}$ , the mass of $A l$ in g [Atomic mass of $A l = 27$ ] used in the reaction is

$2.7$

$54$

$40.5$

$81$

$27$

$4moles 4 A l + 3moles 3 O_{2} ⟶ 2 moles 2 A l_{2} O_{3}$
3 moles of $O_{2}$ combine with 4 moles of $A l$
$∴ 1.5$ moles of $O_{2}$ will combine with $x m o l$ of $A l$
$x = \frac{4}{3} \times 1.5 = 2 m o l$
Thus mass of $A l$ used in the reaction
$= 2 \times 27 g = 54 g$

Q. If 1.5 moles of oxygen combines with Al to form Al2​O3​, the mass of Al in g [Atomic mass of Al=27] used in the reaction is

2.7

54

40.5

81

27

4moles 4Al​+ 3moles 3O2​​⟶2 moles 2Al2​O3​​ 3 moles of O2​ combine with 4 moles of Al ∴1.5 moles of O2​ will combine with xmol of Al x=34​×1.5=2mol Thus mass of Al used in the reaction =2×27g=54g