Question

Calculate the volume of ammonia gas formed at stp when 0.01gof hydrogen gas reacts with nitrogen gas
( N= 14, H=1, M.V=22 4dm^3 at stp

Answer 1

Answer:

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_2A+2HCl→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_2B+2HCl→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