Balance the equation and calculate how many moles of N 2 form when each quantity of reactant completely reacts. N 2 H 4 (l) → NH 3 (g) + N 2 (g) c. 33.9 g N 2 H 4

Alright, so the combustion of octane is represented by the balanced reaction below here, if 39.27 g of propane is combusted, we need to calculate the moles of C. 02 produced. So here we have given propane and we want the amount of moles of C. 02. Okay, so let's go ahead and take a look at this. So we're talking about we have a balanced reaction so we don't need to balance it. Okay? And we're gonna go ahead and start with the number that is given to us. So we have 39.27 g of C eight age 18 of octane is a propane. That is octane, I'm sorry. And we're going to go ahead and convert this into molds by using the molar mass of octane. Okay, so let's go ahead and calculate that. So we have carbon And hydrogen for the carbon, we have eight of them and each of them weigh 12 g. So that is going to give us 96 g for carbon. And then for the hydrogen we have 18, 18 of them times 1.01 oops, times one point oh one g And that will be 18. g. So when we add that together, we get 100 and 14 0. g per mole. Alright, so that's the molar mass of octane. So in one mole of C eight H 18, There are 114.18 g of C8 Age 18. Alright, so we can cancel the g and now we have moles now that we have moles, we can actually do a multiple comparison between octane And c. 0. 2. Okay, so we're gonna put moles of octane on the bottom and from the balanced equation, we can see that we're actually using two moles of octane. So we write two moles of octane. Okay? And then for every two moles of octane, we're producing 16 rolls of CO2. Okay, So this allows us to cancel out moles of octane and get moles of C. 02, and that is exactly what this question is asking. They don't want C 02 and grams, they just wanted to mold. So we're going to go ahead and just leave it right there and find the answer. So let's go ahead and divide and multiply everything here, and we should get 2.75 moles of C 02. Okay, so according to this balanced equation, when we combust, 39.27 g of octane, this is how many moles of C. 02 are going to be produced. Alright, folks, so that is it. That was pretty easy. Right? If you have any questions, let us know and we'll see you in the next video

Ch.4 - Chemical Reactions and Chemical Quantities

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Tro 6th Edition

Ch.4 - Chemical Reactions and Chemical Quantities

Problem 34c

Chapter 4, Problem 34c

Balance the equation and calculate how many moles of N₂ form when each quantity of reactant completely reacts. N₂H₄(l) → NH₃(g) + N₂(g) c. 33.9 g N₂H₄

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Step 1: Write the unbalanced chemical equation: \( \text{N}_2\text{H}_4(l) \rightarrow \text{NH}_3(g) + \text{N}_2(g) \).

Step 2: Balance the chemical equation. Start by balancing the nitrogen atoms. Since there are 2 nitrogen atoms in \( \text{N}_2\text{H}_4 \), you need 2 \( \text{NH}_3 \) molecules to balance the nitrogen: \( \text{N}_2\text{H}_4(l) \rightarrow 2\text{NH}_3(g) + \text{N}_2(g) \).

Step 3: Check the balance of hydrogen atoms. \( \text{N}_2\text{H}_4 \) has 4 hydrogen atoms, and 2 \( \text{NH}_3 \) molecules have 6 hydrogen atoms. Adjust the equation to balance hydrogen: \( 3\text{N}_2\text{H}_4(l) \rightarrow 4\text{NH}_3(g) + 3\text{N}_2(g) \).

Step 4: Calculate the molar mass of \( \text{N}_2\text{H}_4 \) using the atomic masses: \( 2 \times 14.01 + 4 \times 1.01 = 32.05 \text{ g/mol} \).

Step 5: Convert 33.9 g of \( \text{N}_2\text{H}_4 \) to moles using its molar mass: \( \text{moles of } \text{N}_2\text{H}_4 = \frac{33.9 \text{ g}}{32.05 \text{ g/mol}} \). Use the balanced equation to find moles of \( \text{N}_2 \) produced: \( \text{moles of } \text{N}_2 = \frac{3}{3} \times \text{moles of } \text{N}_2\text{H}_4 \).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Balancing Chemical Equations

Balancing chemical equations involves ensuring that the number of atoms of each element is the same on both sides of the equation. This is crucial for obeying the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction. In the given reaction, balancing will help determine the stoichiometric relationships between the reactants and products.

Stoichiometry

Stoichiometry is the calculation of reactants and products in chemical reactions based on balanced equations. It allows chemists to predict how much of a product will form from a given amount of reactant. In this case, stoichiometry will be used to calculate the moles of N2 produced from the given mass of N2H4, using the molar ratios derived from the balanced equation.

Molar Mass

Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It is essential for converting between grams and moles in stoichiometric calculations. For N2H4, the molar mass must be calculated to convert the given mass (33.9 g) into moles, which will then be used to find the amount of N2 produced in the reaction.

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Balance the equation and calculate how many moles of N2 form when each quantity of reactant completely reacts. N2H4(l) → NH3(g) + N2(g) c. 33.9 g N2H4

Verified Solution

Key Concepts

Balancing Chemical Equations

Stoichiometry

Molar Mass

Balance the equation and calculate how many moles of N₂ form when each quantity of reactant completely reacts. N₂H₄(l) → NH₃(g) + N₂(g) c. 33.9 g N₂H₄