1. The document discusses mole to mole and mole to mass conversions for chemical reactions. It provides examples of converting between moles of reactants and products using balanced chemical equations and molar ratios. 2. Specific examples calculate moles of ammonia produced from moles of nitrogen or hydrogen reacted, and moles of nitrogen needed to produce a given amount of ammonia. 3. Other examples calculate grams of oxygen produced from moles of potassium chlorate decomposed, and grams of potassium chlorate needed to produce moles of potassium chloride. The last example calculates moles of potassium chlorate from grams of oxygen produced.

1. Date 5/1/13
Mole to Mole, Mole to Gram
By: Keenan Reardon.
Wednesday, May 1, 13

2. Question 1 Mole to Mole
✤ Start with a balanced equation
✤ N2 + 3 H2 ---> 2 NH3
Wednesday, May 1, 13

3. Question 1
✤ if we have 2.00 mol of N2 reacting, how many moles of NH3 will be
produced?
Wednesday, May 1, 13

4. Process 1
✤ Look at the ratios
✤ N2 + 3 H2 ---> 2 NH3
✤ 2.00 mol of N2-->?molNH3
Wednesday, May 1, 13

5. Answer 2
✤ In the balanced equation for every mole ofnitrogen there were 2
NH3’s so if we have 2 moles of nitrogen then we would need 4
molNH3
✤ =4mol NH3
Wednesday, May 1, 13

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. 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 of
NH3
✤ =4molNH3
Wednesday, May 1, 13

8. Question 3 Mole to Mole
✤ We want to produce 2.75 mol of NH3. How many moles of nitrogen
would be required?
Wednesday, May 1, 13

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, leaving
what we want to be left with on top.
Wednesday, May 1, 13

10. Answer 3
✤ The ratio of the equation is 2/1
✤ We would then cross multiply 2.75/x =2/1
✤ =1.38 mol N2
Wednesday, May 1, 13

11. Question 1 Mole to Mass
✤ Start with balanced equation.
✤ 2 KClO3 ---> 2 KCl + 3 O2
Wednesday, May 1, 13

12. Questions 1
✤ 1.50 mol of KClO3 decomposes. How many grams of O2 will be
produced?
Wednesday, May 1, 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. 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
✤ =72gO2
Wednesday, May 1, 13

15. Question 2
✤ We want to produce 2.75 mol of KCl. How many grams of KClO3
would be required?
✤ 2 KClO3 ---> 2 KCl + 3 O2
Wednesday, May 1, 13

16. Process
✤ Ratios
✤ 2/2
✤ So we know that we are gonna use the same amount of moles, all
we have to do now is convert it to grams
✤ 2 KClO3 ---> 2 KCl + 3 O2
Wednesday, May 1, 13

17. Answer
✤ 2.75mol KClO3 X 122.55g KCLO3
✤ =337g KCLO3
✤ 2 KClO3 ---> 2 KCl + 3 O2
Wednesday, May 1, 13

18. Question 3
✤ If 80.0 grams of O2 was produced, how many moles of KClO3
decomposed?
✤ Different question, but same process, only difference if it is
backwards.
✤ 2 KClO3 ---> 2 KCl + 3 O2
Wednesday, May 1, 13

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 O2
Wednesday, May 1, 13

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 = 2.50molO2
Wednesday, May 1, 13