# Calculating the excess reagent remaining at the end of the reactionhelp?

• miiizpiiink18
In summary, the question is asking about the excess reagent remaining after a reaction between 0.600 moles of Copper (II) Sulfide and 1.40 moles of nitric acid. The equation given is 3 CuS + 8 HNO3 --> 3 Cu(NO3)2 + 3 S + 2 NO + 4 H2. The first part of the question asks how many moles of Copper (II) Nitrate can be produced, with the answer being 0.600 moles for CuS and 0.525 moles for nitric acid. The second part asks for the percent yield, which is found to be 95.2%. Finally, the third
miiizpiiink18
Calculating the excess reagent remaining at the end of the reaction..help?

SO here's the question:

Q: Copper (II) Sulfide (0.600mol) is treated with 1.40mol of nitric acid.

Here's the equation: 3 CuS + 8 HNO3 --> 3 Cu(NO3)2 + 3 S + 2 NO + 4 H2

A) How many moles of Copper (II) Nitrate can be produced?
i already answered this part :D
For CuS, .600 moles of Copper II Nitrate will be produced, and for Nitric acid...0.525 moles of Copper II Nirate will be produced.

B) If 0.500 mol of Copper II Nitrate is actually obtained, what is the percent yield?
for this, I got 95.2% :D

C) Calculate the # of moles of excess reagent remaining at the end of the reaction.
help me with this please. how do i calculate the excess reagent. The excess reagent is CuS, ryyyt =]

miiizpiiink18 said:
SO here's the question:

Q: Copper (II) Sulfide (0.600mol) is treated with 1.40mol of nitric acid.

Here's the equation: 3 CuS + 8 HNO3 --> 3 Cu(NO3)2 + 3 S + 2 NO + 4 H2

A) How many moles of Copper (II) Nitrate can be produced?
For CuS, .600 moles of Copper II Nitrate will be produced, and for Nitric acid...0.525 moles of Copper II Nirate will be produced.

B) If 0.500 mol of Copper II Nitrate is actually obtained, what is the percent yield?
for this, I got 95.2% :D

C) Calculate the # of moles of excess reagent remaining at the end of the reaction.
help me with this please. how do i calculate the excess reagent. The excess reagent is CuS, ryyyt =]

A) Which one is it? 0.6 or 0.525 they cannot be both right.

B) you used the 0.525 so you got it

C) How much does remain after you take 0.525 theoretical (or 0.5 practical value)out of what you had at first?

To calculate the excess reagent remaining at the end of the reaction, we first need to determine the limiting reagent. This is the reactant that is completely used up in the reaction, and therefore determines the amount of product that can be formed.

In this case, we can use the balanced chemical equation to determine the mole ratio between CuS and Cu(NO3)2. We can see that for every 3 moles of CuS, 3 moles of Cu(NO3)2 are produced. Therefore, if we have 0.500 mol of Cu(NO3)2, we must have had 0.500 mol of CuS to begin with.

Next, we can use the given amounts of CuS and nitric acid to determine which is the limiting reagent. Since we have 0.600 mol of CuS and 1.40 mol of nitric acid, nitric acid is in excess. This means that not all of the nitric acid will be used up in the reaction.

To calculate the amount of excess reagent remaining, we can subtract the amount of limiting reagent used from the total amount of excess reagent. In this case, we have used 0.500 mol of CuS, leaving 0.600-0.500 = 0.100 mol of excess CuS remaining at the end of the reaction.

I hope this helps! Let me know if you have any further questions.

## What is an excess reagent?

An excess reagent is a reactant that is present in a greater quantity than is needed for a complete chemical reaction. This means that there will be some of the excess reagent remaining at the end of the reaction.

## Why is it important to calculate the excess reagent remaining at the end of the reaction?

Calculating the excess reagent remaining at the end of the reaction is important because it allows you to determine the actual yield of the reaction. This information is crucial for determining the efficiency of the reaction and for making accurate predictions for future reactions.

## How do you calculate the excess reagent remaining at the end of the reaction?

To calculate the excess reagent remaining at the end of the reaction, you first need to determine the limiting reagent (the reactant that is completely used up). Then, you can use stoichiometry to determine the amount of excess reagent that is left over. This can be done by comparing the moles of the limiting reagent to the moles of the excess reagent in the balanced chemical equation.

## What factors can affect the amount of excess reagent remaining at the end of the reaction?

The amount of excess reagent remaining at the end of the reaction can be affected by factors such as the purity of the reactants, the temperature and pressure of the reaction, and any side reactions that may occur. It is important to carefully control these factors to obtain accurate results.

## Are there any safety precautions to consider when handling excess reagents?

Yes, there are safety precautions to consider when handling excess reagents. Some reagents may be toxic, corrosive, or flammable, so it is important to wear appropriate protective gear and handle them with caution. It is also important to properly dispose of excess reagents to avoid any potential hazards.

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