Answer:i read it Explanation:
The balanced equation :
2NaHCO₃⇒CO₂ + Na₂CO₃+H₂O
<h3>Further explanation</h3>
Given
Reaction
NaHCO(s) --> _CO2+_NaCO(s)+_H2O
Required
The balanced equation
Solution
Maybe the equation should be like this :
NaHCO₃⇒CO₂ + Na₂CO₃+H₂O
Give a coefficient
NaHCO₃⇒aCO₂ + bNa₂CO₃+cH₂O
Make an equation
Na, left=1, right=2b⇒2b=1⇒b=1/2
H, left=1, right=2c⇒2c=1⇒c=1/2
C, left=1, right=a+b⇒a+b=1⇒a+1/2=1⇒a=1/2
The equation becomes :
NaHCO₃⇒1/2CO₂ +1/2Na₂CO₃+1/2H₂O x2
2NaHCO₃⇒CO₂ + Na₂CO₃+H₂O
Answer:
The standard enthalpy change for the reaction at 
is -2043.999kJ
Explanation:
Standard enthalpy change (
) for the given reaction is expressed as:
![\Delta H_{rxn}^{0}=[3mol\times \Delta H_{f}^{0}(CO_{2})_{g}]+[4mol\times \Delta H_{f}^{0}(H_{2}O)_{g}]-[1mol\times \Delta H_{f}^{0}(C_{3}H_{8})_{g}]-[5mol\times \Delta H_{f}^{0}(O_{2})_{g}]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%5E%7B0%7D%3D%5B3mol%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B0%7D%28CO_%7B2%7D%29_%7Bg%7D%5D%2B%5B4mol%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B0%7D%28H_%7B2%7DO%29_%7Bg%7D%5D-%5B1mol%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B0%7D%28C_%7B3%7DH_%7B8%7D%29_%7Bg%7D%5D-%5B5mol%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B0%7D%28O_%7B2%7D%29_%7Bg%7D%5D)
Where 
refers standard enthalpy of formation
Plug in all the given values from literature in the above equation:
![\Delta H_{rxn}^{0}=[3mol\times (-393.509kJ/mol)]+[4mol\times (-241.818kJ/mol)]-[1mol\times (-103.8kJ/mol)]-[5mol\times (0kJ/mol)]=-2043.999kJ](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%5E%7B0%7D%3D%5B3mol%5Ctimes%20%28-393.509kJ%2Fmol%29%5D%2B%5B4mol%5Ctimes%20%28-241.818kJ%2Fmol%29%5D-%5B1mol%5Ctimes%20%28-103.8kJ%2Fmol%29%5D-%5B5mol%5Ctimes%20%280kJ%2Fmol%29%5D%3D-2043.999kJ)
Oxygen because it is on the left of the periodic table so it has a strong pull.
