Question

2 g mixture of two divalent metals A (at wt = 30) and B (at wt = 15) on reacting with dilute HCL solution gives 2.24 L h2 gas at NTP then composition of A (in g)
(1) 1
(2) 0.5
(3) 1.5
(4) 1.2​

Answer 1

answer : option (1) 1

explanation : for metal A, Let mass of metal A is x.

then, no of mole of A = x/30

$A+2HCl\rightarrow ACl_2+H_2$

here it is clear that one mole of metal gives one mole of hydrogen gas.

so, (x/30) mol of metal will give (x/30) mole of hydrogen gas.

similarly, for metal B

mass of metal B = (2 - x)g

then, no of mole of B = (2 - x)/15

$B+2HCl\rightarrow BCl_2+H_2$

here it is clear that one mole of B gives one mole of hydrogen gas.

so, (2 - x)/15 mol of metal will give (2 - x)/15 mole of hydrogen gas.

now, total no of moles of hydrogen gas = x/30 + (2 - x)/15

= (x + 4 - 2x)/30

= (4 - x)/30

given, volume of hydrogen gas at NTP = 2.24 L

so, no of moles of hydrogen gas = 2.24/22.4L = 0.1 mol

so, (4 - x)/30 = 0.1

or, 4 - x = 3

or, x = 1

hence, composition of metal A is 1g

Answer 2

Answer:

1

Explanation:

x/30+2-x/15=0.1,-x+4=3,x=1.