Question

1.5 gm of an element required 2.8 liter of oxygen at s.t.p. for complete combustion. the equivalent weight of element will be
a. 4gm
b. 6 gm
c. 3 gm
d. 8 gm​

Answer 1

Answer:

6g

2.8L O2 for 1.5g element

Then, 22.4L for = 1.5×22.4/2.8 = 12 (at STP 22.4L involves 1 mole)

Equivalent of oxide = Molar mass/2 = 12/2 = 6g

hope this will help you