Answer:
In 33.7 grams SnF2 we have 8.17 grams of F
Explanation:
Step 1: Data given
Mass of SnF2 = 33.7 grams
Molar mass of SnF2 = 156.69 g/mol
Molar mass of F = 19.00 g/mol
Step 2: Calculate moles of SnF2
Moles SnF2 = mass / molar mass
Moles SnF2 = 33.7 grams / 156.69 g/mol
Moles SnF2 = 0.215 moles
Step 3: Calculate moles F
For 1 mol SnF2 we have 2 moles F
For 0.215 moles SnF2 we have 2*0.215 = 0.430 moles F
Step 4: Calculate mass F
Mass F = moles F * molar mass F
Mass F = 0.430 moles * 19.00 g/mol
Mass F = 8.17 grams
In 33.7 grams SnF2 we have 8.17 grams of F
<u>Answer:</u> The amount of energy released per gram of 
is -71.92 kJ
<u>Explanation:</u>
For the given chemical reaction:
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_%7B%28product%29%7D%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28reactant%29%7D%5D)
The equation for the enthalpy change of the above reaction is:
![\Delta H^o_{rxn}=[(5\times \Delta H^o_f_{(B_2O_3(s))})+(9\times \Delta H^o_f_{(H_2O(l))})]-[(2\times \Delta H^o_f_{(B_5H_9(l))})+(12\times \Delta H^o_f_{(O_2(g))})]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%285%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28B_2O_3%28s%29%29%7D%29%2B%289%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28H_2O%28l%29%29%7D%29%5D-%5B%282%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28B_5H_9%28l%29%29%7D%29%2B%2812%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28O_2%28g%29%29%7D%29%5D)
Taking the standard enthalpy of formation:
Putting values in above equation, we get:
![\Delta H^o_{rxn}=[(5\times (1271.94))+(9\times (-285.83))]-[(2\times (73.2))+(12\times (0))]\\\\\Delta H^o_{rxn}=-9078.57kJ](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%285%5Ctimes%20%281271.94%29%29%2B%289%5Ctimes%20%28-285.83%29%29%5D-%5B%282%5Ctimes%20%2873.2%29%29%2B%2812%5Ctimes%20%280%29%29%5D%5C%5C%5C%5C%5CDelta%20H%5Eo_%7Brxn%7D%3D-9078.57kJ)
We know that:
Molar mass of pentaborane -9 = 63.12 g/mol
By Stoichiometry of the reaction:
If 2 moles of produces -9078.57 kJ of energy.
Or,
If 
of produces -9078.57 kJ of energy
Then, 1 gram of will produce = 
of energy.
Hence, the amount of energy released per gram of is -71.92 kJ
The equilibrium constant (Kc) is the product of the equilibrium concentrations of the products raised to their corresponding stoichiometric coefficients divided by the reactants as well. In this case the equilibrium concentration of Cl2 which also applies to SO2 is 1.3x10^-2. The final equilibrium concentration of SO2Cl2 is 9x10^-3. Kc is then equal to 0.0188.
chlorine has the atomic mass of 35.5 and is a non metal in the halogen family
