How much heat is released when a mixture containing 10.0g CS2 and 10.0 g Cl2 reacts by the equation.
CS2 + 3Cl2 ----> S2Cl2 + CCl4
Delta H = -230 kJ
Answered previously. See the original post. I explained proportion there.
To calculate the amount of heat released in this reaction, you will need to use the given equation and the molar masses of the compounds involved.
1. Calculate the number of moles for each compound:
- Moles of CS2 = mass / molar mass = 10.0 g / (12.01 g/mol + 32.07 g/mol) = 0.193 moles
- Moles of Cl2 = mass / molar mass = 10.0 g / (35.45 g/mol + 35.45 g/mol) = 0.141 moles
2. Identify the limiting reactant:
- The stoichiometry of the equation tells us that 1 mole of CS2 reacts with three moles of Cl2. So, we can see that Cl2 is the limiting reactant because we have fewer moles of Cl2 compared to the stoichiometric ratio.
3. Calculate the moles of product formed:
- From the stoichiometry, we know that for every 3 moles of Cl2, we get 1 mole of S2Cl2 and CCl4.
- Therefore, moles of product = moles of limiting reactant (Cl2) / stoichiometric ratio = 0.141 moles / 3 = 0.047 mol
4. Calculate the heat released:
- The given ΔH is -230 kJ, which means it refers to the reaction per mole.
- So, to calculate the heat released for the reaction, we multiply the moles of product formed by ΔH:
Heat released = moles of product * ΔH = 0.047 mol * (-230 kJ/mol) = -10.91 kJ
Therefore, the heat released when the mixture containing 10.0 g CS2 and 10.0 g Cl2 reacts is approximately -10.91 kJ.