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Question

(NH4)2Cr2O7 N2 + Cr2O3 + 4H2O. What volume of nitrogen at STP, will be evolved when 63 g of ammonium dichromate is decomposed? (H = 1, N = 14, O = 16, Cr = 52)

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Solution

Let V be the volume of nitrogen evolved in litres.
Molecular mass of ammonium dichromate = [4(14) + 8(1) + 2(52) + 7(16)] g = 252 g
Number of moles in 63 g of ammonium dichromate = 63252 = 0.25

According to the balanced chemical equation, one mole of dichromate decomposes to give one mole of nitrogen.
Thus, 0.25 mole of dichromate will decompose to give 0.25 mole of nitrogen.

Number of moles of nitrogen at STP = Volume of nitrogenMolar volume= V22.4
V22.4=63252

V = 5.6 L


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