Question

2 N2 H4 (l) + N2 O4 (l) -------> 3 N2 (g) + 4 H2 O(l) ,

if 160 g of N2 H4 is mixed with 160 g N2O4, the yield of H2O is

Answer 1

Answer:

Given the thermochemical equation

N2(g) + O2(g) → 2 NO(g) ΔH = 180.6 kJ

if 558 kJ of energy are supplied, what mass of NO can be made?

Solution

This time, we start with an amount of energy:

558kj

Answer 2

Answer:

6.95 moles of H2O

125.217 grams of H2O

Explanation:

first find limiting reagent

no of moles of N2H4= 160/32 ;

No of moles of N204 =160/92

By observing the denominators we can come a conclusion that

in the reaction that u gave N2O4 IS Limiting reagent

hence

from the equation u can observe that for 1 mole of N2O4 ; about 4 moles of H2O is produced

as u have already seen the no.of moles of N2O4 in the reaction= 160/32

calculating the quantity of H20 :

for 1 mole N2o4 (92 g)----------------------------> 4 moles H2o(4 x 18)

now for 160/92 moles of N2O4( 160 g)----->? moles of H2O( ? grams of N2O2)

cross multiply

Method 1                                                      

finding moles

160 x 4/ 92                                                    

= 6.95 moles of H2O

if wt is asked

we know wt = moles x molecular wt

=6.95 x 18  =125.217 grams of H2O

Method 2

u can directly get wt in single  step of calculation!!

here we are calculating according to the values in the brackets

160 x 4 x 18 / 92 = 11,520/92 = 125.217g of H20 is produced

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