Question

Solid ammonium dichromate decomposes as under:

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

If 63 g of ammonium dichromate decomposes. Calculate;

1. the quantity in moles of (NH₄)₂Cr₂O₇
2. the quantity in moles of nitrogen formed
3. the quantity in moles of nitrogen formed
4. the loss of mass
5. the mass of chromium (III) oxide formed at the same time.

Answer 1

a. Molecular mass of ammonium dichromate = 2(14 + 4) + 104 + 112

= 252 g

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

= `63/252`

= 0.25 moles

b. 1 mole of (NH₄)₂Cr₂O₇ contains nitrogen = 1 mole

∴ 0.25 mole of (NH₄)₂Cr₂O₇ contains nitrogen = 0.25 mole

c. 252 g of ammonium dichromate gives 22.4 dm³ of N₂

63 g of ammonium dichromate gives `(63 xx 22.4)/252`

= 5.6 L

d. 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

e. 252 g of ammonium dichromate gives 152 g of CrO₃

63 g of ammonium dichromate gives `(63 xx 152)/252`

= 38 g