Upcoming SlideShare
×

# Mole to Mole, Mole to Gram

1,464 views

Published on

Published in: Education
1 Like
Statistics
Notes
• Full Name
Comment goes here.

Are you sure you want to Yes No
• Be the first to comment

Views
Total views
1,464
On SlideShare
0
From Embeds
0
Number of Embeds
3
Actions
Shares
0
25
0
Likes
1
Embeds 0
No embeds

No notes for slide

### Mole to Mole, Mole to Gram

1. 1. Date 5/1/13Mole to Mole, Mole to GramBy: Keenan Reardon.Wednesday, May 1, 13
2. 2. Question 1 Mole to Mole✤ Start with a balanced equation✤ N2 + 3 H2 ---> 2 NH3Wednesday, May 1, 13
3. 3. Question 1✤ if we have 2.00 mol of N2 reacting, how many moles of NH3 will beproduced?Wednesday, May 1, 13
4. 4. Process 1✤ Look at the ratios✤ N2 + 3 H2 ---> 2 NH3✤ 2.00 mol of N2-->?molNH3Wednesday, May 1, 13
5. 5. Answer 2✤ In the balanced equation for every mole ofnitrogen there were 2NH3’s so if we have 2 moles of nitrogen then we would need 4molNH3✤ =4mol NH3Wednesday, May 1, 13
6. 6. Question 2 Mole to Mole✤ Suppose 6.00 mol of H2 reacted. How many moles of ammonia(NH3)would be produced?Wednesday, May 1, 13
7. 7. Answer 2✤ This again is about ratios.✤ N2 + 3 H2 ---> 2 NH3✤ The Hydrogen ammonia ratio is 3/2✤ Since we have 6 moles of hydrogen then we would need 4 moles ofNH3✤ =4molNH3Wednesday, May 1, 13
8. 8. Question 3 Mole to Mole✤ We want to produce 2.75 mol of NH3. How many moles of nitrogenwould be required?Wednesday, May 1, 13
9. 9. Process✤ This is tricky so we start with what is given to us.✤ Which is 2.75 mol of NH3--> ?MolN2✤ We then would have to use the coefﬁcients and cross multiply, leavingwhat we want to be left with on top.Wednesday, May 1, 13
10. 10. Answer 3✤ The ratio of the equation is 2/1✤ We would then cross multiply 2.75/x =2/1✤ =1.38 mol N2Wednesday, May 1, 13
11. 11. Question 1 Mole to Mass✤ Start with balanced equation.✤ 2 KClO3 ---> 2 KCl + 3 O2Wednesday, May 1, 13
12. 12. Questions 1✤ 1.50 mol of KClO3 decomposes. How many grams of O2 will beproduced?Wednesday, May 1, 13
13. 13. Process✤ 2 KClO3 ---> 2 KCl + 3 O2✤ Find ratios, KClO3 and O2 is 2/3✤ 1.50 mol of KClO3--> ?grams O2✤ Cross multiply - 1.50/x times 2/3✤ =2.25mol O2✤Wednesday, May 1, 13
14. 14. Answer 1✤ Now that we have 2.25mol O2 we convert it to grams.✤ We then multiply 2.25mol O2 by the molar mass of one mole of O2✤ 2.25mol O2 X 32g O2✤ =72gO2Wednesday, May 1, 13
15. 15. Question 2✤ We want to produce 2.75 mol of KCl. How many grams of KClO3would be required?✤ 2 KClO3 ---> 2 KCl + 3 O2Wednesday, May 1, 13
16. 16. Process✤ Ratios✤ 2/2✤ So we know that we are gonna use the same amount of moles, allwe have to do now is convert it to grams✤ 2 KClO3 ---> 2 KCl + 3 O2Wednesday, May 1, 13
17. 17. Answer✤ 2.75mol KClO3 X 122.55g KCLO3✤ =337g KCLO3✤ 2 KClO3 ---> 2 KCl + 3 O2Wednesday, May 1, 13
18. 18. Question 3✤ If 80.0 grams of O2 was produced, how many moles of KClO3decomposed?✤ Different question, but same process, only difference if it isbackwards.✤ 2 KClO3 ---> 2 KCl + 3 O2Wednesday, May 1, 13
19. 19. Process✤ Find Ratio✤ O2 to KCLO3 is 3/2✤ we then divide 80.0 grams of O2 by the molar mass of O2✤ 80/32 = 2.50molO2✤ 2 KClO3 ---> 2 KCl + 3 O2Wednesday, May 1, 13
20. 20. Answer 3✤ Now we cross multiply 2.50/x = 3/2✤ This equals 1.67 molKClO3✤ 2 KClO3 ---> 2 KCl + 3 O2✤ 80/32 = 2.50molO2Wednesday, May 1, 13