Limiting Reactant Tutorial By Riley Nabozny Jaylene Lesher To navigate through the tutorial, just click the mouse button!

HERE’S A QUESTION… If there are 10 bowls of ice cream and 4 cherries to make ice cream sundaes, how many ice cream sundaes can be made??

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

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.

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

Quick Review of Balancing Equations 0 Unbalanced equation:- H 2 SO 4 + Fe ---> Fe 2 (SO 4 ) 3 + H 2 0 Balance the SO 4 first (as it is a complex ion and it is in one chemical species on each side) 0 3H 2 SO 4 + Fe ---> Fe 2 (SO 4 ) 3 + H 2 0 Now balance the Fe (which is also in one chemical on each side) 0 3H 2 SO 4 + 2Fe ---> Fe 2 (SO 4 ) 3 + H 2 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:- 3H 2 SO 4 + 2Fe ---> Fe 2 (SO 4 ) 3 + 3H 2 Go back to: 1. Balance the equation1. Balance the equation

Steps for a limiting reactant problem 1.Balance the equation. 2.Find the number of moles each reactant using the mass of reactant giving. 3.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.) 4.Multiply the product’s number of moles by its molecular weights to find mass produced. 5.Find excess reactant and the amount of left over reactant. **ALWAYS USE SIG FIGS!!**

Example Question Suppose that a solution containing 4.50 grams of Na 3 PO 4 is mixed with a solution containing 7.40 grams of Ba(NO 3 ) 2. How many grams of Ba 3 (PO 4 ) 2 can be formed and which is the limiting reactant?

1. Balance the equation Na 3 PO 4 + Ba(NO 3 ) 2 → Ba 3 (PO 4 ) 2 + NaNO 3 Becomes 2 Na 3 PO Ba(NO 3 ) 2 → 1 Ba 3 (PO 4 ) NaNO 3 through balancing the reactants and products. Review of Balancing Equations

2. Find the number of moles of each reactant using the mass of reactant given and their molecular weights g 7.40 g 2 Na 3 PO Ba(NO 3 ) 2 → 1 Ba 3 (PO 4 ) NaNO m m (4.50 grams Na 3 PO 4 ) x ( ) = moles Na 3 PO 4 (7.40 grams Ba(NO 3 ) 2 ) x ( ) = moles Ba(NO 3 ) 2

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* Na 3 PO 4 : Ba 3 (PO 4 ) 2 Ba(NO 3 ) 2 : Ba 3 (PO 4 ) 2 2:13:1 2 Na 3 PO Ba(NO 3 ) 2 → 1 Ba 3 (PO 4 ) NaNO m m moles Na 3 PO 4 x ( ) = moles Ba 3 (PO 4 ) moles Ba(NO 3 ) 2 x ( ) = moles Ba 3 (PO 4 ) 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

4. Multiply the product’s number of moles by its molecular weight to find mass produced NOTE: Use 3 sig figs! moles Ba 3 (PO 4 ) 2 x ( ) = 8.25 grams moles Ba 3 (PO 4 ) x ( ) = 5.66 grams Ba(NO 3 ) 2 is the limiting reactant! Return to Question 1Return to Question 2

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

5. To find amount of excess reactant, calculate how much non-limiting reactant (Na 3 PO 4 ) actually reacted with the limiting reactant grams of Ba(NO 3 ) 2 x ( ) x ( ) x ( ) = 3.09 grams Na 3 PO 4 reacted Return to Question 5

6. Take the amount reacted and subtract it from original amount to determine amount left over grams – 3.09 grams = 4.31 grams Na 3 PO 4 remaining Return to Question 5

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

If 2.35 grams of H 2 gas react with grams of N 2 gas to make ammonia gas (NH 3 )… 1) How many grams of NH 3 can form with the reactant H 2 ? A)A) 30.2 g NH 3 B)B) 26.7 g NH 3 C)C) 20.3 g NH 3 D)D) g NH 3

If 2.35 grams of H 2 gas react with grams of N 2 gas to make ammonia gas (NH 3 )… 2) How many grams of NH 3 can you make with the reactant N 2 ? A)A) 98.4 g NH 3 B)B) 26.7 g NH 3 C)C) 183 g NH 3 D)D) 54.0 g NH 3

If 2.35 grams of H 2 gas react with grams of N 2 gas to make ammonia gas (NH 3 )… 3) Which reactant is the ‘limiting reactant’? A)A) H 2 B)B) N 2 C)C) Both reactants D)D) Neither reactants

If 2.35 grams of H 2 gas react with grams of N 2 gas to make ammonia gas (NH 3 )… 4) Which reactant is the ‘excess reactant’? A)A) H 2 B)B) N 2 C)C) Both reactants D)D) Neither reactants

If 2.35 grams of H 2 gas react with grams of N 2 gas to make ammonia gas (NH 3 )… 5) How much excess reactant is left over? A)A) 0 grams B)B) 128 grams C)C) 34.5 grams D)D) 78.2 grams

Oops, That’s incorrect. You can…. 1)Return to Question 1Return to Question 1 2)See explanationSee explanation

Oops, That’s incorrect. You can… 1.Return to Question 2Return to Question 2 2.See explanationSee explanation

Oops, That’s incorrect. You can… 1.Return to Question 3Return to Question 3 2.See explanationSee explanation

Oops, That’s incorrect. You can… 1.Return to Question 4Return to Question 4 2.See explanationSee explanation

Oops, That’s incorrect. You can… 1.Return to Question 5Return to Question 5 2.See explanationSee explanation

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