At one time, a common means of forming small quantities of oxygen gas in the laboratory was to heat KClO₃: 2 KClO₃(s) → 2 KCl(s) + 3 O₂(g) ΔH = -89.4 kJ For this reaction, calculate H for the formation of (b) 10.4 g of KCl.

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hey everyone in this example, we have the following reaction And we need to find the entropy of formation of 5.15 g of lead chloride. So we want to recognize whether our equation is balanced and based on our psyche aama tree and the coefficients. It's definitely a balanced equation. Our next step is to recognize that we are given the entropy of formation for reaction for our formation of lead chloride as negative 3 36 point okay, joules per mole and this is for one mole of our lead chloride. So we want to recognize that that comes from the coefficient one which is in front of our product lead chloride. So our next step is to first because we are given the amount in grams for lead chloride. We need to convert from grams to moles. So we want to start from 5.15 g of our lead chloride. And then we're going to go from grams to moles by recalling our molar mass for the compound. So when we find lead in group four a on our periodic tables, we see that it has a molar mass of 207.2 g per mole. And when we find chlorine in group seven a on the periodic table, We see that it has a molar mass of 35.4, 5 g per mole. Now recognize that in our formula, we have two moles of our Chlorine Atom. And so we're going to multiply this by two. So to get the total mass for the compound lead chloride, we're going to have these two masses added up together And we're going to have a total of 278.1 g per mole for our compound P BCL two. So we're going to plug that in. So we have to 78.1 g for one mole of our lead chloride. So this allows us to cancel out our units of grams leaving us with moles, which is what we want. And we're going to get a value equal to 0.01815 moles of our lead chloride. And actually I re I just need to rewrite this number. So it's actually 0.018 519 moles of our lead chloride. And so now we can calculate our anthem P A formation for our lead chloride For our 5.15 g of lead chloride. And we would get that that is equal to our entropy of formation for one mole, which is given as negative 3 36.0 Killer Jewels for one mole of lead chloride. And then we're multiplying this by our molds of our 5.15 g of our lead chloride which we found above. So we plugged that in as 0. moles of our lead chloride. So this allows us to cancel out most, leaving us with kayla jewels as our final unit, Which is definitely what we need for entropy of formation. And we get a value in our calculators equal to negative 6.22238. So we can round this to about 366 so that we have negative 6.22 kg joules as our final answer For the entropy of formation of our 5. g of our product, lead chloride. So I hope that everything we reviewed was clear. If you have any questions, please leave them down below and I will see everyone in the next practice video.

At one time, a common means of forming small quantities of oxygen gas in the laboratory was to heat KClO₃: 2 KClO₃(s) → 2 KCl(s) + 3 O₂(g) ΔH = -89.4 kJ For this reaction, calculate H for the formation of (b) 10.4 g of KCl.

Verified Solution

Key Concepts

Stoichiometry

Enthalpy Change (ΔH)

Molar Mass

At one time, a common means of forming small quantities of oxygen gas in the laboratory was to heat KClO3: 2 KClO3(s) → 2 KCl(s) + 3 O2(g) ΔH = -89.4 kJ For this reaction, calculate H for the formation of (b) 10.4 g of KCl.

Verified Solution

Key Concepts

Stoichiometry

Enthalpy Change (ΔH)

Molar Mass

At one time, a common means of forming small quantities of oxygen gas in the laboratory was to heat KClO₃: 2 KClO₃(s) → 2 KCl(s) + 3 O₂(g) ΔH = -89.4 kJ For this reaction, calculate H for the formation of (b) 10.4 g of KCl.