Question

Name: Date: Per: 7) U + 3 Br? ? UBr? What is my actual yield of uranium hexabromide if I start with 100 grams of uranium and get a percent yield of 83% ? 8) H?SO? ? H?O + SO? If I start with 89 grams of sulfuric acid and produce 7.1 grams of water, what is my percent yield? Limiting Reagent Worksheet Using your knowledge of stoichiometry and limiting reagents, answer the following questions: 1) Write the balanced equation for the reaction of lead (II) nitrate with sodium iodide to form sodium nitrate and lead (II) iodide: 2) If I start with 25.0 grams of lead (II) nitrate and 15.0 grams of sodium iodide, how many grams of sodium nitrate can be formed? 3) What is the limiting reagent in the reaction described in problem 2? 4) How much of the non limiting reagent will be left over from the reaction in problem #2?

          Name:
Date:
Per:
7) U + 3 Br? ? UBr?
What is my actual yield of uranium hexabromide if I start with 100 grams of uranium and get a
percent yield of 83% ?
8) H?SO? ? H?O + SO?
If I start with 89 grams of sulfuric acid and produce 7.1 grams of water, what is my percent
yield?
Limiting Reagent Worksheet
Using your knowledge of stoichiometry and limiting reagents, answer the following questions:
1) Write the balanced equation for the reaction of lead (II) nitrate with sodium iodide to form
sodium nitrate and lead (II) iodide:
2) If I start with 25.0 grams of lead (II) nitrate and 15.0 grams of sodium iodide, how many grams
of sodium nitrate can be formed?
3) What is the limiting reagent in the reaction described in problem 2?
4) How much of the non limiting reagent will be left over from the reaction in problem #2?

Show more…

Name:
Date:
Per:
7) U + 3 Br? ? UBr?
What is my actual yield of uranium hexabromide if I start with 100 grams of uranium and get a
percent yield of 83% ?
8) H?SO? ? H?O + SO?
If I start with 89 grams of sulfuric acid and produce 7.1 grams of water, what is my percent
yield?
Limiting Reagent Worksheet
Using your knowledge of stoichiometry and limiting reagents, answer the following questions:
1) Write the balanced equation for the reaction of lead (II) nitrate with sodium iodide to form
sodium nitrate and lead (II) iodide:
2) If I start with 25.0 grams of lead (II) nitrate and 15.0 grams of sodium iodide, how many grams
of sodium nitrate can be formed?
3) What is the limiting reagent in the reaction described in problem 2?
4) How much of the non limiting reagent will be left over from the reaction in problem #2?

Added by Omar Y.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert Madhur L

06/09/2023

Step 1

Step 1: The balanced equation for the reaction of lead (II) nitrate with sodium iodide to form sodium nitrate and lead (II) iodide is: Pb(NO₃)₂ + 2NaI → PbI₂ + 2NaNO₃ Show more…

Show all steps

lock

Thanks for your feedback!

U + 3 Br₂ → UBr₆ What is my actual yield of uranium hexabromide if I start with 100 grams of uranium and get a percent yield of 83% ? H₂SO₄ → H₂O + SO₂ If I start with 89 grams of sulfuric acid and produce 7.1 grams of water, what is my percent yield? Limiting Reagent Worksheet Using your knowledge of stoichiometry and limiting reagents, answer the following questions: Write the balanced equation for the reaction of lead (II) nitrate with sodium iodide to form sodium nitrate and lead (II) iodide: If I start with 25.0 grams of lead (II) nitrate and 15.0 grams of sodium iodide, how many grams of sodium nitrate can be formed? What is the limiting reagent in the reaction described in problem 2? How much of the non limiting reagent will be left over from the reaction in problem #2?

Play audio

Feedback

Powered by NumerAI

Madhur L and 65 other subject Chemistry 101 educators are ready to help you.

Ask a new question

Labs

Want to see this concept in action?

NEW

Explore this concept interactively to see how it behaves as you change inputs.

View Labs

Key Concepts

Recommended Videos

Recommended Textbooks

Transcript

00:01 Hello students, so in this video we have a reaction where lead nitrate reacts with sodium iodide to produce sodium nitrate plus lead iodide.

00:16 Now we have 25 g of lead nitrate and 15 g of sodium iodide.

00:22 So firstly to find the number of moles we have to find the molecular masses of the pbno3 that is equals to 207 .2 plus 14 plus 48 into 2 that is equals to 331 .2 and the mass of nai is equals to 23 plus 127 that is equals to 150 g.

00:51 Now let us find the number of moles that is 25 by 331 .2 that is equals to 0 .075 and 15 by 150 that is equals to 0 .1.

01:04 Now we will find the limiting reagent.

01:07 So here you can see that they are reacting in the ratio 1 is to 2 and this is more than this but it is required double of this.

01:16 So if we will write that 1 mole of lead nitrate is equivalent to 2 moles of nai.

01:27 So we can write that 0 .1 mole of nai which we have will be 0 .1 by 2 moles of lead nitrate which that is 0 .05 moles.

01:45 So we have 0 .05 moles actually we have 0 .075 moles.

01:49 So this means that nai is getting completely react is reacting completely and some amount of lead nitrate is left.

01:58 So we can say that here nai is the limiting reagent.

02:05 Now we will find the amount of nano3 produced.

02:09 So from here you can see that 2 moles of nai is producing 2 moles of nano3.

02:19 So that means that if we cut it, it will be 1 mole each.

02:23 So we can say that nai the weight was there is we have 0 .1 mole of nai.

02:33 So that is equivalent to 0 .1 mole of nano3...

Ace is your personal tutor. It breaks down any question with clear steps so you can learn.

Start Using Ace

Continue solving this problem

Create a free account to:

View full step-by-step solution
Ask follow-up questions with Ace AI
Save progress and study later