Question

Solid ammonium dichromate decomposes as under:

\[\ce{(NH4)2Cr2O7 ->N2 + Cr2O3 + 4H2O}\]

If 126 g of ammonium dichromate decomposes, calculate:

1. the number of moles of ammonium dichromate that undergoes decomposition.
2. the mass of chromic oxide formed at the same time.
3. the volume of nitrogen gas evolved at STP.

[At. wt: N = 14, Cr  = 52, O = 16, H = 1]

Answer 1

Given: \[\ce{(NH4)2Cr2O7 ->N2 + Cr2O3 + 4H2O}\]

Mass = 126 g

At. wt: N = 14, Cr  = 52, O = 16, H = 1

Molar Mass of (NH₄)₂Cr₂O₇:

= 2(14) + 4(1) + 2(52) + 7(16)

= 36 + 104 + 112 

= 252 g

Molar Mass of Cr₂O₃:

= 2(52) + 3(16) 

= 104 + 48 

= 152 g

a. 252 g of ammonium dichromate is 1 mole. 

∴ 126 g of ammonium dichromate = `1/252 xx 126`

= `1/2`

= 0.5 moles

b. 252 g of ammonium dichromate gives 152 g of chromic oxide.

∴ 126 g of ammonium dichromate provides = `152 × 126/252`

= `152/2`

= 76 g of chromic oxide

c. 252 g of ammonium dichromate produces 22.4 L of N₂ at STP. 

∴ 126 g of ammonium dichromate produces = `22.4 × 126/252`

= `22.4/2`

= 11.2 L of N₂ at STP