Answer: Enthalpy of combustion (per mole) of 
is -2657.5 kJ
Explanation:
The chemical equation for the combustion of butane follows:
The equation for the enthalpy change of the above reaction is:
![\Delta H^o_{rxn}=[(8\times \Delta H^o_f_{CO_2(g)})+(10\times \Delta H^o_f_{H_2O(g)})]-[(1\times \Delta H^o_f_{C_4H_{10}(g)})+(4\times \Delta H^o_f_{O_2(g)})]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%288%5Ctimes%20%5CDelta%20H%5Eo_f_%7BCO_2%28g%29%7D%29%2B%2810%5Ctimes%20%5CDelta%20H%5Eo_f_%7BH_2O%28g%29%7D%29%5D-%5B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7BC_4H_%7B10%7D%28g%29%7D%29%2B%284%5Ctimes%20%5CDelta%20H%5Eo_f_%7BO_2%28g%29%7D%29%5D)
We are given:
Putting values in above equation, we get:
![\Delta H^o_{rxn}=[(8\times -393.5)+(10\times -241.82)]-[(2\times -125.6)+(4\times 0)]\\\\\Delta H^o_{rxn}=-5315kJ](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%288%5Ctimes%20-393.5%29%2B%2810%5Ctimes%20-241.82%29%5D-%5B%282%5Ctimes%20-125.6%29%2B%284%5Ctimes%200%29%5D%5C%5C%5C%5C%5CDelta%20H%5Eo_%7Brxn%7D%3D-5315kJ)
Enthalpy of combustion (per mole) of is -2657.5 kJ
Answer:
84.8%
Explanation:
Step 1: Given data
Bob measured out 1.60 g of Na. He forms NaCl according to the following equation.
Na + 1/2 Cl₂ ⇒ NaCl
According to this equation, he calculates that 1.60 g of sodium should produce 4.07 g of NaCl, which is the theoretical yield. However, he carries out the experiment and only makes 3.45 g of NaCl, which is the real yield.
Step 2: Calculate the percent yield.
We will use the following expression.
%yield = real yield / theoretical yield × 100%
%yield = 3.45 g / 4.07 g × 100% = 84.8%
The answer to the molar mass of Ca(NO3)2 is gonna be C. 164.1 g/mol
