3.9 g of a mixture of aluminium and its oxide on reaction with aqueous solution of sodium hydroxide, gave 840 mL of a gas under standard conditions. Thus, aluminium content in the mixture is

Question

3.9 g of a mixture of aluminium and its oxide on reaction with aqueous solution of sodium hydroxide, gave 840 mL of a gas under standard conditions. Thus, aluminium content in the mixture is (A) 3.225 g (B) 0.675 g (C) 1.35 g (D) 2.70 g

Tardigrade Admin · Accepted Answer

Al + NaOH + H 2 O arrow NaAlO 2+1.5 H 2 <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/f0e2ddd35832de2994beca536c9b1bbf-.png" /> 33.6 L H 2 gas are from =27 g Al 0.840 L H 2 gas are from =(27 × 0.840/33.6) g Al =0.675 g Al

Q. $3.9 g$ of a mixture of aluminium and its oxide on reaction with aqueous solution of sodium hydroxide, gave $840 m L$ of a gas under standard conditions. Thus, aluminium content in the mixture is

$3.225 g$

$0.675 g$

$1.35 g$

$2.70 g$

$A l + N a O H + H_{2} O \to N a A l O_{2} + 1.5 H_{2}$

$33.6 L H_{2}$ gas are from $= 27 g A l$
$0.840 L H_{2}$ gas are from $= \frac{27 \times 0.840}{33.6} g A l$
$= 0.675 g A l$

Q. 3.9g of a mixture of aluminium and its oxide on reaction with aqueous solution of sodium hydroxide, gave 840mL of a gas under standard conditions. Thus, aluminium content in the mixture is

3.225g

0.675g

1.35g

2.70g