Will Rate Lifesaver ASAP Consider the reaction N 2 (g) + 3H 2 (g)<====> 2NH 3 (g

Question

Will Rate Lifesaver ASAP
Consider the reaction N2(g) + 3H2(g)<====> 2NH3(g). You inject 250ml ofN2(g) and 750ml of
H2(g), both at STP into an uninflated balloon. Assume that STP remains constant both inside and outside theballoon. Calculate the volume if the reaction goes to 30%completion. (Hint: Avogadro's law and stoichiometry ofreacting volumes) Will Rate Lifesaver ASAP
Consider the reaction N2(g) + 3H2(g)<====> 2NH3(g). You inject 250ml ofN2(g) and 750ml of
H2(g), both at STP into an uninflated balloon. Assume that STP remains constant both inside and outside theballoon. Calculate the volume if the reaction goes to 30%completion. (Hint: Avogadro's law and stoichiometry ofreacting volumes)

Explanation / Answer

We Know that :      N2(g) + 3H2(g)<====> 2NH3(g)          According togram molar volume 1 mole of any gas at STP condition occupies22,400 mL volume     Number of moles of N2 taken = 250 ml / 22,400 mL                                                       = 0.011 moles     Number of moles of H2taken = 750 mL / 22400 mL                                                    = 0.033 moles       As the number of molescalculated obeys the stochiometric equation :              N2(g) + 3H2(g) <====>2NH3(g)        I : 0.011     0.033                  0   ---- Intial number of moles when reaction hasn'tstarted        C:  0.0033   0.0297             0.6    ---- reaction for 30 % completion        E:   0.0077    0.0033            0.6     --- number of moles atequilibrium       Total number of moles atequilibrium :          ( 0.0077+ 0.0033 +0.6 ) moles       = 0.611 moles         The volume of theballoon after 30% completion of the reaction -           V = 0.611 moles x 0.0821 atm-L / mol-K x 273 K / 1 atm                = 13.69 L      N2(g) + 3H2(g)<====> 2NH3(g)          According togram molar volume 1 mole of any gas at STP condition occupies22,400 mL volume     Number of moles of N2 taken = 250 ml / 22,400 mL                                                       = 0.011 moles     Number of moles of H2taken = 750 mL / 22400 mL                                                    = 0.033 moles       As the number of molescalculated obeys the stochiometric equation :              N2(g) + 3H2(g) <====>2NH3(g)        I : 0.011     0.033                  0   ---- Intial number of moles when reaction hasn'tstarted        C:  0.0033   0.0297             0.6    ---- reaction for 30 % completion        E:   0.0077    0.0033            0.6     --- number of moles atequilibrium       Total number of moles atequilibrium :          ( 0.0077+ 0.0033 +0.6 ) moles       = 0.611 moles         The volume of theballoon after 30% completion of the reaction -           V = 0.611 moles x 0.0821 atm-L / mol-K x 273 K / 1 atm                = 13.69 L    

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