Question

Answer the questions completely with complete calculations.

1. what is the oxidation number of elements in BrO2-, BrO3- & BrO4-, Br2, Br-?

2. In the reaction 4 NH3 (g) + 5 O2 (g) --> 4 NO (g) + 6 H2O (l)

750 g NH3 is mixed with 750 g of O2.

a. what is the amount of NO produced in this reaction?

b. Which is the limiting reagent?

c. How much of the excess reagent remains?

d. The actual yield of NO obtained was 400g. What is the % yield of NO?

3. A compound of 49.4% K, 20.3% S, and 30.3% O by mass. What is the empirical formula of the compound?

Answer 1

1. Oxidation Nos Of Br in different species -

BrO2^- = +3

calculations ; x + (-4 ) = -1    x = +3 ,where x represents oxidation number of Br , and O.N.

of O is = (-2 )

Similarly oxidation numbers of element Br in other given species are calcuulated as in,

BrO3^-   = +5

BrO4^-   = +7  

Br₂   = 0 .......[ it is as per convention for di atomic - gases ]

Br ^- = -1.........[ oxidation number of halides is always -1, as per convention ]

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2.

Study the stoichiometry of the given reaction-

....................................4 NH3 (g) + 5O2(g) -----------> 4NO(g) + 6H2 O(l)

.....................................4moles................5 moles..................4moles..............6 moles

hence the theretical ratio of moles of NH3 to O2 = 4 / 5 = 0.80

Given amounts as moles-

.................................... 750 / 17..............750 / 32...........................................................

.....................................=44.118................=23.44

.......................... ratio of the moles of NH3 to O2 taken for reaction = 44.118 / 23 44

..............................................................................................................= 1.88

Now , we know that limiting reagent is-

a reagent which gets consumed first, and therefore limits the amounts of products formed.

Thus we find that O 2(g) has been taken in lesser amounts ,than what is required for the reaction.

Therefore it will get consumed first.Hence-

b) O2 is the limiting reagent in this case.

.a) Amount of NO produced in this case , therefore , should depend upon the moles of O2 taken

for reaction as NH3 has been taken in excess.

As`per stoichiometry , 5 moles of O2 when react with NH3 produce 4 moles of NO

........................... 23.44 moles.........................................( 4 x 23.44 ) / 5..............

......................................................................................................= 18.752 moles......

.............................................................................................or, = 18.752 x 30 = 562.56 gms.

c) Calculations for excess reagent left;

Again , as per stoichiometry,

....................5 moles of O2(g) react with 4 moles of NH3(g)

...............23.44.....................................(4 x 23.44 ) / 5...........

.............................................................= 18.752 moles.............

hence , moles of excess reagent ,NH3(g ) left = ( 44.113 - 18.752 )

.........................................................................= 25.361 moles

....................................................................or, = 25.361 x 17

...........................................................................= 431.14 gms.

d) Calculation for percentage yield of NO(g)

.........% yield = (400 x 100) / 562.56

....................... = 71.10 %

3 )

Empirical formula of the given compound is........K₂ S O₂

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