Limiting Reactant

1. LimitingReactant Tutorial By Riley Nabozny JayleneLesher To navigate through the tutorial, just click the mouse button!

2. If there are 10 bowls of ice cream and 4 cherries to make ice cream sundaes, how many ice cream sundaes can be made?? Here’s a question…

3. The Answer is… 4 Sundaes!.... ….6 bowls of ice cream left over!

4. The cherries are what in the recipe??... That’s a lot of left over ice cream! Because there is a limited amount of cherries, only a small number of sundaes can be yielded.

5. Limiting Reactants By definition : The reactant in a chemical reaction that limits the amount of product that can be formed. The reaction will stop when all of the limiting reactant is consumed. Return to Question 3

6. Now that we’ve had our dessert first, time to get to work on these problems!

7. Quick Review of Balancing Equations • Unbalanced equation:- H2SO4 + Fe ---> Fe2(SO4)3 + H2 • Balance the SO4 first (as it is a complex ion and it is in one chemical species on each side) • 3H2SO4 + Fe ---> Fe2(SO4)3 + H2 • Now balance the Fe (which is also in one chemical on each side) • 3H2SO4 + 2Fe ---> Fe2(SO4)3 + H2 Finally, balance the hydrogen (although it is in one chemical species on each side, it is usually a good idea to leave it until last) Balanced Equation:- 3H2SO4 + 2Fe ---> Fe2(SO4)3 + 3H2 Go back to: 1. Balance the equation

8. Steps for a limiting reactant problem • Balance the equation. • Find the number of moles each reactant using the mass of reactant giving. • Determine the moles of product by looking at the mole ratio; Multiply of divide the products by their mole ratios. (Does not matter which product is tested first.) • Multiply the product’s number of moles by its molecular weights to find mass produced. • Find excess reactant and the amount of left over reactant. **ALWAYS USE SIG FIGS!!**

9. Example Question Suppose that a solution containing 4.50 grams of Na3PO4 is mixed with a solution containing 7.40 grams of Ba(NO3)2. How many grams of Ba3(PO4)2 can be formed and which is the limiting reactant?

10. 1. Balance the equation Na3PO4 + Ba(NO3)2 → Ba3(PO4)2 + NaNO3 Becomes 2 Na3PO4 + 3Ba(NO3)2 → 1 Ba3(PO4)2 + 6 NaNO3 through balancing the reactants and products. Review of Balancing Equations

11. 2. Find the number of moles of each reactant using the mass of reactant given and their molecular weights. 4.50 g 7.40 g 2 Na3PO4 + 3 Ba(NO3)2 → 1 Ba3(PO4)2 + 6 NaNO3 0.0274 m0.0283 m (4.50 grams Na3PO4) x ( ) = 0.0274 moles Na3PO4 (7.40 grams Ba(NO3)2 ) x ( ) = 0.0283 moles Ba(NO3)2

12. 3. Determine the moles of product by looking at the mole ratio: Multiply or divide the products by their mole ratios NOTE: It does not matter which product is tested first *Include 4 sig figs* Na3PO4: Ba3(PO4)2Ba(NO3)2 : Ba3(PO4)2 2:13:1 2Na3PO4 + 3 Ba(NO3)2 → 1 Ba3(PO4)2 + 6 NaNO3 0.0274 m0.0283 m 0.0274 moles Na3PO4 x ( ) = 0.0137 moles Ba3(PO4)2 0.0283 moles Ba(NO3)2 x ( ) = 0.0094 moles Ba3(PO4)2 *You can see here which reactant is the limiting reactant: but to determine the mass yielded, continue to the next step. Return to Question 1Return to Question 2

13. 4. Multiply the product’s number of moles by its molecular weight to find mass produced NOTE: Use 3 sig figs! 0.0137 moles Ba3(PO4)2 x ( ) = 8.25 grams 0.0094 moles Ba3(PO4) x ( ) = 5.66 grams Ba(NO3)2 is the limiting reactant! Return to Question 1Return to Question 2

14. Excess Reactant Definition: The reactant in a chemical reaction that remains when a reaction stops. The excess reactant remains because there is nothing with which it can react with. Return to Question 4

15. 5. To find amount of excess reactant, calculate how much non-limiting reactant (Na3PO4) actually reacted with the limiting reactant. 7.40 grams of Ba(NO3)2 x ( ) x ( ) x ( ) = 3.09 grams Na3PO4 reacted Return to Question 5

16. 6. Take the amount reacted and subtract it from original amount to determine amount left over. 7.40 grams – 3.09 grams = 4.31 grams Na3PO4 remaining Return to Question 5

17. Let’s take a test! NOTE: Click on the letter that you think is the correct answer!

18. If 2.35 grams of H2 gas react with 0.190 grams of N2 gas to make ammonia gas (NH3)… 1) How many grams of NH3 can formwith the reactant H2? A) 30.2 g NH3 B) 26.7 g NH3 C) 20.3 g NH3 D) 24. 8 g NH3

19. If 2.35 grams of H2 gas react with 0.190 grams of N2 gas to make ammonia gas (NH3)… 2) How many grams of NH3 can you make with the reactant N2? A) 98.4 g NH3 B) 26.7 g NH3 C) 183 g NH3 D) 54.0 g NH3

20. If 2.35 grams of H2 gas react with 0.190 grams of N2 gas to make ammonia gas (NH3)… 3) Which reactant is the ‘limiting reactant’? A) H2 B) N2 C)Both reactants D)Neither reactants

21. If 2.35 grams of H2 gas react with 0.190 grams of N2 gas to make ammonia gas (NH3)… 4) Which reactant is the ‘excess reactant’? A) H2 B) N2 C) Both reactants D) Neither reactants

22. If 2.35 grams of H2 gas react with 0.190 grams of N2 gas to make ammonia gas (NH3)… 5) How much excess reactant is left over? A) 0 grams B) 128 grams C) 34.5 grams D) 78.2 grams

23. Oops, That’s incorrect. You can…. Return to Question 1 See explanation

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33. Thank you for participating in our tutorial! Go celebrate with a bowl of ice cream! Congratulations!