A 1.073 gram mixture of hydrogen and nitrogen is contained in asample volume of
ID: 685289 • Letter: A
Question
A 1.073 gram mixture of hydrogen and nitrogen is contained in asample volume of 750.0 cm3 at298K and 3.318 bar initially. Assuming ideal gas behavior,calculate the mole fractions and
partial pressures of the gases (i)initially and (ii)after the twogases react to form the maximum
quantity of ammonia.
Explanation / Answer
n = PV/RT = [(3.318 bar * 105 Pa/1bar) *(750 cm3 *(1m/100 cm)3]/ (8.314 J/mol/K * 298 K) n = 0.1004 moles of H2 and N2 Let x = mol fraction of H2 Then 1 - x = mol fraction of N2 x*n = moles of H2 (1-x)*n = moles of N2 H2 has molar mass of 2g/mol N2 hasmolar mass of 28 g/mol 1.073 gram = x*n *(2 g/mol) + (1-x)*n *(28 g/mol) 1.073 = (2x - 28x)*n + 28*n x = (1.073 - 28*n)/(-26n) = (1.073 - 28*0.1004) / (-26*0.1004) x = 0.66604 mol fraction of H2 is 0.666 (about 66.6 %) 1 - x = mol fraction N2 = 0.333956934 ~ 0.334 ==================== 3H2 + N2 => 2NH3 identify the limiting reagent initial moles N2 = 0.334*0.1004 moles = 0.03354 moles N2 moles of H2 to react with N2 = 0.03354 moles N2 * (3 moles H2/ 1mol N2) = 0.1006 moles H2 required But there is 0.666*0.1004 = 0.06689 moles H2 available So H2 is limiting reactant; all H2 reacts 0.06689 moles H2 *( 1 mol N2 / 3 mol H2) = 0.022298 moles N2reacted moles N2 left = 0.03354 moles N2 - 0.022298 moles N2 reacted =0.011245 moles N2 left moles NH3 formed = 0.06689 moles H2 *( 2 mol NH3 / 3 mol H2) =0.044596 moles NH3 formed 0 moles H2 left total gas moles=0.011245 moles N2 + 0.044596 moles NH3 = 0.05584moles total gas mol frac N2 = 0.011245/0.05584 = 0.20137699 ~ 0.201 mol frac NH3 = 0.044596/0.05584 = 0.79862301 ~ 0.799 mol frac H2 = 0
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