Mixture of two bivalent metals A and B having mass 2 g when dissolved in HCl at STP, 2.24 L H2 is evolved. What is the mass of A present in mixture? (Atomic mass of A = 15 u, B = 30 u)

Question

Mixture of two bivalent metals A and B having mass 2 g when dissolved in HCl at STP, 2.24 L H2 is evolved. What is the mass of A present in mixture? (Atomic mass of A = 15 u, B = 30 u) (A) 1 g (B) 1.5 g (C) 0.5 g (D) 0.75 g

Tardigrade Admin · Accepted Answer

A +2 HCl longrightarrow ACl 2+ H 2 Mole: (x/15) (x/15) B +2 HCl longrightarrow BCl 2+ H 2 Mole: (2-x/30) (2-x/30) Total moles of H 2=(x/15)+(2-x/30)=(2.24/22.4)=(1/10) ⇒ x=1 g

Q. Mixture of two bivalent metals $A$ and $B$ having mass $2 g$ when dissolved in $H Cl$ at STP, $2.24 L H_{2}$ is evolved. What is the mass of A present in mixture? (Atomic mass of $A = 15 u, B = 30 u$ )

1 g

1.5 g

0.5 g

0.75 g

$A + 2 H Cl ⟶ A C l_{2} + H_{2}$
Mole $: \frac{x}{15} \frac{x}{15}$
$B + 2 H Cl ⟶ BC l_{2} + H_{2}$
Mole $: \frac{2 - x}{30} \frac{2 - x}{30}$
Total moles of $H_{2} = \frac{x}{15} + \frac{2 - x}{30} = \frac{2.24}{22.4} = \frac{1}{10}$
$\Rightarrow x = 1 g$

Q. Mixture of two bivalent metals A and B having mass 2g when dissolved in HCl at STP, 2.24LH2​ is evolved. What is the mass of A present in mixture? (Atomic mass of A=15u,B=30u)

1 g

1.5 g

0.5 g

0.75 g

A+2HCl⟶ACl2​+H2​ Mole :15x​15x​ B+2HCl⟶BCl2​+H2​ Mole :302−x​302−x​ Total moles of H2​=15x​+302−x​=22.42.24​=101​ ⇒x=1g

Q. Mixture of two bivalent metals $A$ and $B$ having mass $2 g$ when dissolved in $H Cl$ at STP, $2.24 L H_{2}$ is evolved. What is the mass of A present in mixture? (Atomic mass of $A = 15 u, B = 30 u$ )

$A + 2 H Cl ⟶ A C l_{2} + H_{2}$
Mole $: \frac{x}{15} \frac{x}{15}$
$B + 2 H Cl ⟶ BC l_{2} + H_{2}$
Mole $: \frac{2 - x}{30} \frac{2 - x}{30}$
Total moles of $H_{2} = \frac{x}{15} + \frac{2 - x}{30} = \frac{2.24}{22.4} = \frac{1}{10}$
$\Rightarrow x = 1 g$