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

[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 =]

 

[Lok]

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 =]

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?

 

[Blueshift5]

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.