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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.
Solution 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
Solution 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
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