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At 340K 1atm pressure N2O4 is 66% dissociated into NO2.what is volume of 10g of N2O4 occupy under these conditions
N2O4(g) 2NO2(g) 10.0g T = 340K P = 1.00atm Assuming we started with 10.0g of N2O4 and, 66% of it decomposed, that will leave 3.4g of N2O4, and make 6.6g of NO2. Convert those to moles, and then plug them into the ideal gas equation. At equilibrium.... 3.4 g N2O4 .... 0.03696 mol N2O4 ...... carry extra sig figs for intermediate answers and 6.6g NO2 ....... 0.1435 mol NO2 Total moles = 0.1805 moles of N2O4 and NO2, at a total pressure of 1.00 atm. Find the volume. PV = nRT V = nRT / P = 0.1805 mol x 0.0821Latm/molK x 340K / 1.00 atm = 5.04L You will get 5.02L if you use 0.18 moles of gas
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