Solid ammonium dichromate decomposes as the equationr:
(NH44)2Cr2O7→N2+Cr2O3+4H2O
If 63 g of sammonium dichromate decomposes, calculate:
(a) the quantity in moles of (NH4)2Cr2O7
(b) the qunatity in moles of nitrogen formed
(c) the volume of N2 evolved at STP
(d) the loss of mass
(e) the mass of chromium (iii) oxide formed at the same time.
(NH44)2Cr2O7→N2+Cr2O3+4H2O
(a) Molecular mass of (NH44)2Cr2O7=252g
Mole=given massmolar mass=63252=0.25
(b) 1 mole (NH44)2Cr2O7 produces 1 mole of N2
0.25 mole (NH44)2Cr2O7 produces 0.25 mole of N2
(c) 1 mole N2 occupies 22.4 litre
So, 0.25 mole of N2 will occupy = 0.25×22.4=5.6 litres
(d)
252 g of solid ammonium dichromate decomposes to give 152 g of solid chromium oxide, so the loss in mass in terms of solid formed = 100 g
Now, if 63 g ammonium dichromate is decomposed, the loss in mass would be =100×63252=25g
(e) If 252 g of ammonium dichromate produces Cr2O3=152g
So, 63 g ammonium dichromate will produce =63×152252=38g