Consider the reaction:P4 (s)+ 6 Cl2 (g) = 4PCl3(g)
Delt H = -1230 kJ
How much heat is released when 52.5 of Cl2 (g) reacts with an excess of P4 (s)?
I tried the question and got -910.8 kJ and the answer is suppose to be -/+ 152 kJ.
Moles Cl2=52.5/70.9=.74 moles
heat released= .74/6*1230=-152 kJ
but....0.74/6*1230= 0.74/7380=0.000100271 moles..
(0.74/6)1230=152kj
To determine the heat released in this reaction, you can use stoichiometry and the given value of ΔH (enthalpy change).
First, calculate the moles of Cl2 reacting by using the given mass and the molar mass of Cl2.
Molar mass of Cl2 = 2 * 35.45 g/mol = 70.90 g/mol
Moles of Cl2 = mass of Cl2 / molar mass of Cl2
Moles of Cl2 = 52.5 g / 70.90 g/mol
Next, using the balanced equation, determine the ratio between Cl2 and heat released.
From the balanced equation: P4 (s) + 6 Cl2 (g) → 4 PCl3 (g)
The coefficient in front of Cl2 is 6, which means that for every 6 moles of Cl2 that react, 1230 kJ of heat is released.
So, the heat released for 1 mole of Cl2 is:
Heat released per mole of Cl2 = -1230 kJ / 6 mol = -205 kJ/mol
Finally, calculate the heat released for the given mass of Cl2:
Heat released = Heat released per mole of Cl2 * Moles of Cl2
Heat released = -205 kJ/mol * (moles of Cl2 obtained earlier)
Now, let's calculate the result using the given data:
Moles of Cl2 = 52.5 g / 70.90 g/mol = 0.74 mol
Heat released = -205 kJ/mol * 0.74 mol = -151.7 kJ
Therefore, the heat released when 52.5 g of Cl2 reacts with excess P4 is approximately -151.7 kJ, which is close to the expected answer of -/+ 152 kJ. It seems that the difference in values might be due to rounding errors or calculation discrepancies.