I'm pretty sure its Sankey Diagram.
(Edit: Turned Dankey to Sankey which is the proper name for the diagram)
First find the oxidation states of the various atoms:
<span>in Cr2O2 2- Cr @ +1; In NH3 N @ +3; in CrO3 Cr @ +3, N2 N @ 0 </span>
<span>Note that N gained electrons, ie, was reduced; Cr was oxidized </span>
<span>Now there is a problem, because B has NH4+ which the problem did not, and is not balanced, showing e- in/out </span>
<span>B.NH4+ → N2 </span>
<span>Which of the following is an oxidation half-reaction? </span>
<span>A.Sn 2+ →Sn 4+ + 2e- </span>
<span>Sn lost electrons so it got oxidized</span>
Answer : The enthalpy of combustion per mole of
is -2815.8 kJ/mol
Explanation :
Enthalpy change : It is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as 
The equation used to calculate enthalpy change is of a reaction is:
![\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%5Eo_f%28product%29%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%5Eo_f%28reactant%29%5D)
The equilibrium reaction follows:

The equation for the enthalpy change of the above reaction is:
![\Delta H^o_{rxn}=[(n_{(CO_2)}\times \Delta H^o_f_{(CO_2)})+(n_{(H_2O)}\times \Delta H^o_f_{(H_2O)})]-[(n_{(C_6H_{12}O_6)}\times \Delta H^o_f_{(C_6H_{12}O_6)})+(n_{(O_2)}\times \Delta H^o_f_{(O_2)})]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%28n_%7B%28CO_2%29%7D%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28CO_2%29%7D%29%2B%28n_%7B%28H_2O%29%7D%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28H_2O%29%7D%29%5D-%5B%28n_%7B%28C_6H_%7B12%7DO_6%29%7D%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28C_6H_%7B12%7DO_6%29%7D%29%2B%28n_%7B%28O_2%29%7D%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28O_2%29%7D%29%5D)
We are given:

Putting values in above equation, we get:
![\Delta H^o_{rxn}=[(6\times -393.5)+(6\times -285.8)]-[(1\times -1260)+(6\times 0)]=-2815.8kJ/mol](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%286%5Ctimes%20-393.5%29%2B%286%5Ctimes%20-285.8%29%5D-%5B%281%5Ctimes%20-1260%29%2B%286%5Ctimes%200%29%5D%3D-2815.8kJ%2Fmol)
Therefore, the enthalpy of combustion per mole of
is -2815.8 kJ/mol
The mass of 1 mole of an element is its atomic weight on the Periodic Table in grams.
1g = 1000mg
The mass of 1 mole of Kr = 83.80g Kr
Convert mg Kr to g Kr.
398mg Kr x (1g/1000mg) = 0.398g Kr
Convert g Kr to mol Kr.
0.398g Kr x (1mol Kr/83.80g Kr) = 4.75x10-3mol Kr