Question

Solid ammonium dichromate decomposes as:

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

If 63 g of ammonium dichromate decomposes. Calculate what will be the loss of mass.

Answer 1

Number of moles = `"Weight"/"Molecular weight"`

= `63/252`

= 0.25 moles

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

0.25 moles of ammonium dichromate gives 0.25 moles of N₂ = 7 g

 1 mole of H₂O = 18 g

∴ Total loss of mass = 7 + 18

= 25 g

Answer 2

\[\ce{\underset{252}{[NH4]2Cr2O7} -> \underset{litre}{\underset{22.4}{N2}}↑ + \underset{152}{Cr2O3} + 4H2O}\]

Above 100°C, nitrogen and water vapours will escape leaving only Cr₂O₃.

252 g substance loses = (252 − 152) = 100 g

∴ 63 g substance loses = `100/252 xx 63`

= 25 g