Question

solid ammonium dichromate decomposes as under:
(NH4)2Cr2O7---->N2+Cr2O3+4H2O
If 63 g of ammonium dichromate decomposes,calculate:
a)the quantity in moles of (NH4)2Cr2O7
b)the quantity in moles of nitrogen formed
c)the volume of N2 evolved at STP
d)the loss of mass
e)the mass of chromium(III) oxide formed at the same time.

Answer 1

Answer:

(NH_4)_2Cr_2O_7 ---------gt N_2 + Cr_2O_3 + 4H_2O

According to above equation, 1 mol of ammonium dichromate would give 4 mols of H_2O . When heated these H_2O will be evaporated away.

The molar mass of ammonium dichromate

= 2 xx(14+4)+ 2xx52 + 7 xx 16

= 252 g per mol

The number of mols of ammonium dichromate heated

= 63 g / 252 g per mol

= 0.25 mol.

Therefore number of mols of water evaporated = 4 xx 0.25 mol

= 1 mol.

The mass of water evaporated = 1 mol xx 18 g per mol

= 18 g.

Therefore 18 grams of water is evaporated.

Also produced N_2 also are released to atmosphere.

Number of N_2 mols produced = 1 xx 0.25 mol

= 0.25 mol

Mass of N_2 released = (14+14) xx 0.25 g

= 7 g.

Therefore total mass loss = 18 + 7

= 25 g.

Answer 2

Answer:

I) 0.25 mole

ii) 0.25 mole

iii) 5.6 litres

iv) 25 g

v) 38 g