Question #425b9

1 Answer
Jarni Renz
Jan 27, 2018

#etaN_2=0.624mol; mN_2=17.5g#
#etaH_2O=1.25mol; mH_2O=22.5g#

Explanation:

  1. Write and balance the equation
    #N_2H_4(l)+O_2(g)->N_2(g)+2H_2O(g)#
  2. Find the molar masses of the two reactants. Refer to the periodic table for the relative atomic masses of elements composing them.
    #N_2H_4=(32.05g)/(mol)#
    #O_2=(32.00g)/(mol)#
  3. Given the masses of the reactants, find individual number of moles.
    #ul(etaN_2H_4):#
    #=20.0cancel(gN_2H_4)xx(1molN_2H_4)/(32.06cancel(gN_2H_4))=0.624molN_2H_4#
    #ul(etaO_2):#
    #=20.0cancel(gO_2)xx(1molO_2)/(32.00cancel(gO_2))=0.625molO_2#
  4. Now, find the limiting reactant by multiplying each involved compound to its molar ratio; i.e.,
    #color(red)ul(etaN_2H_4 " available"=0.624mol):#
    #=0.624cancel(molN_2H_4)xx(1molO_2)/(1cancel(molN_2H_4))=0.624molO_2#

    This means that

#0.624molN_2H_4-=0.624molO_2#, but

#(etaO_2 " available")/(0.625molO_2)>(etaO_2 " required")/(0.624molO_2)#

#color(blue)ul(etaO_2 " available"=0.625mol):#
#=0.625cancel(molO_2)xx(1molN_2H_4)/(1cancel(molO_2))=0.625molN_2H_4#
This means that

#0.625molO_2-=0.625molN_2H_4#

#(etaN_2H_4" available")/(0.624molN_2H_4)<(etaN_2H_4 " required")/(0.625molN_2H_4)#

Therefore; #color(red)(etaN_2H_4 " is the limiting reactant")#

5.Now, find the products in grams using the limiting reactant and the mole ratios of the involved compounds as described in the balanced equation above; i.e.,

#ul(N_2 " produced")#
#=0.624cancel(molN_2H_4)xx(1molN_2)/(1cancel(molN_2H_4))=0.624molN_2, then#
#=0.624cancel(molN_2)xx(28.02gN_2)/(1cancel(molN_2))=17.5gN_2#
#ul(H_2O " produced")#
#=0.624cancel(molN_2H_4)xx(2cancel(molH_2O))/(1cancel(molN_2H_4))=1.25molH_2O#
#=1.25molH_2Oxx(18gH_2O)/(1cancel(molH_2O))=22.5gH_2O#

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