When 10.00 g of phosphorus is burned in O₂(g) to form P₄O₁₀(s), enough heat is generated to raise the temperature of 2950 g of water from 18.0 °C to 38.0 °C. Calculate the enthalpy of formation of P₄O₁₀(s) under these conditions.

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hey everyone, we're told that solid iron reacts with oxygen gas to form iron oxide. And the reaction from 269.3 g of iron generates enough heat to raise the temperature of 91.4 kg of water, which was initially at 15.5°C to 20.2°C Were asked to calculate the standard entropy of formation of our iron to three oxide. In order to answer this question, we need to recall the relationship that the heat of our reaction is equivalent to the negative heat of our calorie emitter. And we know that the queue of our reaction is going to be our N times our delta H. F of our reaction. And in this case our delta HF, which is our standard entropy information is going to be our iron to three oxide and our negative Q of our calorie emitter is going to be the negative of our mass. Times of specific heat times the change in temperature in this case it's going to be in regard to water. So, plugging this all in, we get the delta H. F of our iron to three oxide is going to be equivalent to the negative mass of our water. Times the specific heat of water times the change in temperature of our water divided by n. Now let's go ahead and plug in our values. So we have negative 91.4 kg of our water and we want to convert this into grams. So we know that one kg is equivalent to 10 to the third grams. We're going to multiply this by the specific heat of water, which is 4.184 jewels over mole times degrees Celsius. And we have our change in temperature which is 20.2°C -15.5°C. Now let's go ahead and divide this by our end. So we had 269.3 g of our iron and we want to use our molar mass of iron. So we know that we have 55.85 g of iron for one mole of iron. And this is where our dimensional analysis comes in. And looking at our multiple ratios, we know that we have three mole of iron per one mole of our iron. 2, 3 oxide, calculating this out, We end up with a value of negative 1118.3 kg joules per mole. And using the correct number of significant figures, we simplify this to negative 1.12 times 10 to the third. Killer joules per mole. And this is going to be our final answer. So I hope that made sense. And let us know if you have any questions

When 10.00 g of phosphorus is burned in O₂(g) to form P₄O₁₀(s), enough heat is generated to raise the temperature of 2950 g of water from 18.0 °C to 38.0 °C. Calculate the enthalpy of formation of P₄O₁₀(s) under these conditions.

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Key Concepts

Enthalpy of Formation

Calorimetry

Stoichiometry

When 10.00 g of phosphorus is burned in O2(g) to form P4O10(s), enough heat is generated to raise the temperature of 2950 g of water from 18.0 °C to 38.0 °C. Calculate the enthalpy of formation of P4O10(s) under these conditions.

Verified Solution

Key Concepts

Enthalpy of Formation

Calorimetry

Stoichiometry

When 10.00 g of phosphorus is burned in O₂(g) to form P₄O₁₀(s), enough heat is generated to raise the temperature of 2950 g of water from 18.0 °C to 38.0 °C. Calculate the enthalpy of formation of P₄O₁₀(s) under these conditions.