Question

Question #425b9

Answer 1

`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`