N2 (g) + 3 CL2 (g) ---> 2 NCL3 (g)
delta H = + 230 kj
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a) What mass of N2 would absorb 96.5 kj of heat as it reacts?
b) What volume of Cl2 at STP would be required to react with the N2 in part (a)?
I will be happy to critique your thinking.
I don't know how to find mass with the enthalpy formula.
Just rephrase the question.
If 230 kJ is delta H for 1 mol of N2, then how many mols would account for96.5 kJ
Thanks.
To calculate the mass of N2 that would absorb 96.5 kJ of heat, we need to use the equation:
delta H = q/m
where delta H is the heat absorbed (given as +230 kJ), q is the heat absorbed (96.5 kJ), and m is the mass of N2.
Rearranging the equation to solve for m, we have:
m = q / delta H
Substituting the given values, we get:
m = 96.5 kJ / 230 kJ
m = 0.42 grams (rounded to two decimal places)
Therefore, the mass of N2 that would absorb 96.5 kJ of heat is approximately 0.42 grams.
To calculate the volume of Cl2 at STP (Standard Temperature and Pressure) required to react with the N2 from part (a), we need to use the balanced equation:
N2 (g) + 3 Cl2 (g) ---> 2 NCl3 (g)
From the balanced equation, we can see that the stoichiometric ratio between N2 and Cl2 is 1:3. This means that for every 1 mole of N2, we need 3 moles of Cl2.
To find the moles of N2, we can use the formula:
moles = mass / molar mass
The molar mass of N2 is 28 g/mol.
Substituting the given mass of N2 from part (a):
moles of N2 = 0.42 grams / 28 g/mol
moles of N2 = 0.015 moles (rounded to three decimal places)
Since the stoichiometric ratio between N2 and Cl2 is 1:3, we need three times the number of moles of Cl2. Therefore, the number of moles of Cl2 required is:
moles of Cl2 = 3 * moles of N2
moles of Cl2 = 3 * 0.015 moles
moles of Cl2 = 0.045 moles
Now, we need to convert the moles of Cl2 to volume at STP.
At STP, 1 mole of any gas occupies 22.4 liters. Therefore, the volume of Cl2 at STP can be calculated as:
volume = moles * 22.4 L/mol
Substituting the given moles of Cl2:
volume = 0.045 moles * 22.4 L/mol
volume = 1.008 liters (rounded to three decimal places)
So, the volume of Cl2 at STP required to react with the N2 from part (a) is approximately 1.008 liters.