Question

1.80 gm of a metal oxide required 833ml of hydrogen at ntp to be reduced to its metal .find the equivalent weight of oxide and metal

Answer 1

M^+2 
MO + H2 --> M + H2O 
M2O + H2 --> 2 M + H2O 
1.8 g MO / 0.03719 mole = 48.4 g/mol of MO or M2O 
M = 32.4 g/mol 

M^+4 
MO2 + 2 H2 --> M + 2 H2O 
1.8 g MO2 / 0.01860 mole = 96.8 g/mol 
M = 64.8 g/mol 

M^+3 
M2O3 + 3 H2 --> 2 M + 3 H2O 
1.8 g M2O3 / 0.01240 = 145 
M = 48.5 g/mol 

M^+1 
M2O + H2 --> 2 M + H2O 
1.8 g M2O / 0.03719 = 48.4 
M = 24.2 g/mol 

Titanium(III)oxide 
Ti2O3 is fairly close. Titanium is 47.9 g/mol vs 48.5 calculated value

Answer 2

Here is the answer to this question

Explanation:

Let’s suppose that metal oxide to be MxOy .

The reaction of hydrogen gas with metal oxide is:

MxOy + y H2 →→ xM + yH2O

The Molar mass of this H2 gas = 2g/mol

At STP

Mass of 22.4 L of H2 gas = 1 mole

Therefore, 833 mL of H2 gas = 122.4×8331000122.4×8331000 g = 0.0372 moles H2

Now the ‘n’ factor (total number of electrons gained or lost during a reaction) for H2 is 2

Therefore, the gram equivalent of H2 = Number of moles ××n

= 0.0371×× 2

= 0.0742 g eq

Gram equivalent of H2 = Gram equivalent of metal oxide

Gram equivalent = Given mass equivalent mass Given mass equivalent mass

Thus equivalent mass of metal oxide = 1.800.0721.800.072= 25

An equal metal oxide weight contain equal mass of oxygen and metal (8).

Hence proved

Equivalent mass of metal = 25 - 8 = 17

Hence equivalent mass of metal is 17.