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Solid ammonium dichromate decomposes as under:(NH4)2Cr2O7---->N2+Cr2O3+4H2OIf 63 g of ammonium dichromate decomposes,calculate:a)the quantity in moles of (NH4)2Cr2O7b)the quantity in moles of nitrogen formedc)the volume of N2 evolved at STPd)the loss of masse)the mass of chromium(III) oxide formed at the same time. |
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Answer» Answer: (NH_4)_2Cr_2O_7 ---------gt N_2 + Cr_2O_3 + 4H_2O According to above equation, 1 mol of ammonium dichromate would give 4 mols of H_2O . When heated these H_2O will be evaporated away. The molar mass of ammonium dichromate = 2 XX(14+4)+ 2xx52 + 7 xx 16 = 252 g per mol The number of mols of ammonium dichromate heated = 63 g / 252 g per mol = 0.25 mol. Therefore number of mols of water evaporated = 4 xx 0.25 mol = 1 mol. The mass of water evaporated = 1 mol xx 18 g per mol = 18 g.
Therefore 18 grams of water is evaporated. Also produced N_2 also are released to atmosphere. Number of N_2 mols produced = 1 xx 0.25 mol = 0.25 mol Mass of N_2 released = (14+14) xx 0.25 g = 7 g.
Therefore TOTAL mass loss = 18 + 7 = 25 g. |
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