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

Answer 1

Answer: (B)

0.675 g

Since $A{l}_2{O}_3$ does not react with $NaOH$, the chemical reaction may be represented as

$2Al+2NaOH+6H_{2}O\to 2NaAl(OH{)}_4+3H_2$

At STP, $840mL$ of gas$=\frac{0.84}{22.4}=0.0375moles$

Since 3 moles of gas is liberated by 2 moles of aluminium,

0.0375 moles of gas is liberated by $\frac{2}{3}\times 0.0375=0.035$ moles of aluminium

Hence, mass of aluminium$=0.035\times 27=0.675g$

Hence, B is the correct answer