Respuesta :
Answer:
-3.72 (or -3.70 depending on what values you used)
Explanation:
First, use the molar mass of Cl2 convert the mass of Cl2 to moles.
1.48 g Cl2(1 mol70.906 g)=0.02087 mol Cl2
Note that we are given ΔH=−886kJ. This refers to the enthalpy change associated with the reaction of 5mol of Cl2 by the balanced equation shown below.
2P+5Cl2⟶2PCl5ΔH=−886kJ
Therefore, to determine the enthalpy change associated with the reaction of 1.48gCl2, divide ΔH by 5molCl2 to determine the enthalpy change per mole of Cl2, then multiply by 0.02087 mol Cl2. (note: if you round up here to .021 mol of Cl2 you will get the final answer of -3.72 later)
0.02087 mol Cl2(−886 kJ5 mol Cl2)=−3.698 kJ
Rounding the answer should to three significant figures, we find that the enthalpy change associated with the reaction of 1.48gCl2 is −3.70 kJ.
Notice that coefficients in stoichiometric equations (indicating numbers of moles) are exact, so they do not constrain the number of significant figures.
The enthalpy change for the reaction when 1.48 g of Cl₂ reacts with excess P is –3.69 KJ.
We'll begin by calculating the number of mole of in 1.48 g of Cl₂. This can be obtained as follow:
Molar mass of Cl₂ = 2 × 35.5 = 71 g/mol
Mass of Cl₂ = 1.48 g
Mole of Cl₂ =?
Mole = mass / molar mass
Mole of Cl₂ = 1.48 / 71
Mole of Cl₂ = 0.0208 mole
Finally, we shall determine the enthalpy change for the reaction. This can be obtained as follow:
2P + 5Cl₂ —> 2PCl₂ ΔH = −886 KJ
From the balanced equation above,
5 moles of Cl₂ reacted to produce an enthalpy change of −886 KJ.
Therefore,
0.0208 mole of Cl₂ will react to produce an enthalpy change of = [tex]\frac{0.0208 * - 886}{5}\\\\[/tex] = –3.69 KJ.
Thus, the enthalpy change for the reaction when 1.48 g of Cl₂ reacts with excess P is –3.69 KJ.
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