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प्रश्न
Solid ammonium dichromate decomposes as under:
\[\ce{(NH4)2Cr2O7 -> N2 + Cr2O3 + 4H2O}\]
If 63 g of ammonium dichromate decomposes. Calculate;
- the quantity in moles of (NH4)2Cr2O7
- the quantity in moles of nitrogen formed
- the quantity in moles of nitrogen formed
- the loss of mass
- the mass of chromium (III) oxide formed at the same time.
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उत्तर
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
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