4HCl + O₂ → 2Cl₂ + 2H₂O
mole ratio of HCl : O₂ is 4 : 1
∴ if moles of HCl = 2.3 mol
then mol of O₂ = 2.3 mol ÷ 4
= 0.575 mol
mass of O₂ = moles of O₂ × molar mass of O₂
= 0.575 mol × (16 × 2) g/mol
= 18.4 g
It is a type of applied science, concretely speaking, Physics applied.
Answer: The enthalpy of the given reaction is 1234.8kJ/mol.
Explanation: Enthalpy change of the reaction is the amount of heat released or absorbed in a given chemical reaction.
Mathematically,

For the given reaction:





![\Delta H_{rxn}=\Delta H_f_{(C_2H_5OH)}+3\Delta H_f_{(O_2)}-[2\Delta H_f_{(CO_2)}+3\Delta H_f_{(H_2O)}]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5CDelta%20H_f_%7B%28C_2H_5OH%29%7D%2B3%5CDelta%20H_f_%7B%28O_2%29%7D-%5B2%5CDelta%20H_f_%7B%28CO_2%29%7D%2B3%5CDelta%20H_f_%7B%28H_2O%29%7D%5D)
Putting values in above equation, we get:
![\Delta H_{rxn}=[-277.6+3(0)]-[2(-393.5)+3(-241.8)]kJ/mol](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5B-277.6%2B3%280%29%5D-%5B2%28-393.5%29%2B3%28-241.8%29%5DkJ%2Fmol)

<span>BaCl2+Na2SO4---->BaSO4+2NaCl
There is 1.0g of BaCl2 and 1.0g of Na2SO4, which is the limiting reagent?
"First convert grams into moles"
1.0g BaCl2 * (1 mol BaCl2 / 208.2g BaCl2) = 4.8 x 10^-3 mol BaCl2
1.0g Na2SO4 * (1 mol Na2SO4 / 142.04g Na2SO4) = 7.0 x 10^-3 mol Na2SO4
(7.0 x 10^-3 mol Na2SO4 / 4.8 x 10^-3 mol BaCl2 ) = 1.5 mol Na2SO4 / mol BaCl2
"From this ratio compare it to the equation, BaCl2+Na2SO4---->BaSO4+2NaCl"
The equation shows that for every mol of BaCl2 requires 1 mol of Na2SO4. But we found that there is 1.5 mol of Na2SO4 per mol of BaCl2. Therefore, BaCl2 is the limiting reagent.</span>
Answer:
Kinetic energy because it has motion, Potential energy bc there is still more distance to fall and heat energy bc there is heat from friction.
Explanation: