How to Determine the Limiting Reagent in a Chemical Reaction
Introduction
This is a set of instructions that will teach high school or undergraduate chemistry students how to determine the limiting reagent in a chemical reaction. Finding the limiting reagent in a chemical reaction will show which element or compound will run out first and be limited. These instructions will help to determine how much product will be produced in the chemical reaction. Any high school or undergraduate chemistry student with basic knowledge of chemistry and stoichiometry will be able to complete these instructions.
Materials
• Pencil with an eraser
• At least one sheet of ruled paper
• Math calculator (ex. TI-30XA or TI-84)
Warning
Be sure to include all
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Write down the following problem using a pencil and a sheet of paper: (See Figure 1)
Titanium (IV) chloride is obtained from Titanium (IV) oxide from the following reaction:
3TiO2 (s) + 4C (s) + 6Cl2 (g) ↔ 3TiCl4 (g) + 2CO2 (g) + 2CO (g)
Using the reaction above, determine the limiting reagent when 9.00 grams of TiO2 (s) are mixed with 13.00 grams of Cl2 (g) and excess carbon. Given: Atomic Mass of Reactants • TiO2 (s) =76.860 grams • Cl2 (g) =70.906 grams Figure 1: Limiting Reagent Problem
2. Set up a train track with the starting amount of TiO2(s) found in the original problem. (See Figure 2) Figure 2: Starting amount of TiO2 (s) on Train Track
3. Write the atomic mass for TiO2 (s) found in the original problem on the bottom train track, to the right of the vertical line. (See Figure 3) Figure 3: Atomic Mass of TiO2 (s)
4. Write 1 mole TiO2 (s) on the top train track, to the right of the vertical line. (See Figure 4) 1 mole of anything is equal to the amount of its atomic mass. 1 mole of TiO2 (s) is equal to its atomic mass given in the original problem. Figure 4: 1 mole of TiO2 (s) on Train
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Write the number determined from Step 19, including units, on the bottom train track, next to the vertical line drawn in Step 18. (See Figure 14) Hint: the units are moles of Cl2 (g). Figure 14: Amount of Moles of Cl2 (g)
21. Repeat Step 8.
22. Write the number determined from Step 21, along with units, on the top train track, next to the vertical line drawn in Step 18. (See Figure 15) Hint: the units are moles of TiCl4 (g). Figure 15: Amount of Moles of TiCl4 (g)
23. Repeat Steps 11-13.
24. Write the solution from Step 23 next to the fraction, labeling with correct units. (See Figure 16) Hint: the units are moles of TiCl4 (g).This is how many moles of TiCl4 (g) were produced from Cl2 (g). Figure 16: Solution after Simplifying the Fraction
25. Compare the final number of moles of TiCl4 (g) produced from both TiO2 (s) and Cl2 (g). (See Figure 17) Figure 17: Determining the Limiting Reagent
26. Determine which number of moles is smaller. Whichever compound, TiO2(s) or Cl2 (g), produced the lowest number of moles of TiCl4 (g), is the limiting reagent.
Limiting Reagent= Cl2 (g)
Troubleshooting
Question: What happens if both solutions are the same amount?
Answer: Double check your numbers and recalculate your
Number of moles of〖 K_2 CrO〗_4, mol = (4.0 ×〖10〗^(-2) mol L^(-1))(5.0×〖10〗^(-2) L)= 2.0×〖10〗^(-3) mol
We then took 1ml of the 0.1% solution from test tube 2 using the glucose pipette and added it to test tube 3, we then used the H2O pipette and added 9ml of H2O into test tube 3 creating 10ml of 0.01% solution.
The purpose of this lab was to calculate the percent composition by mass of oxygen in potassium chlorate.
The mass of Mg + the mass of O2=mass of MgxOx. Knowing the mass of
the Ar for Oxygen is 16, I can find the number of moles for Magnesium
At the top of the first lift hill (a), there is maximum potential energy because the train is as high as it gets. As the train
Determining the Relative Atomic Mass of Lithium An experiment has been carried out to determine the relative atomic mass of Lithium by using two different types of methods The first method that was carried out was to determine the volume of Hydrogen produced. In this experiment a fixed amount of Lithium was used, in my case it was 0.11g. At the end of this experiment, the volume of Hydrogen gas I collected was 185cm³. Then using the solution of lithium hydroxide made from experiment one, I used it in the titrating experiment, to find out the total volume of Hydrochloric acid used to titrate the lithium hydroxide. RESULTS TABLE Experiment Initial Volume ( cm³) Final Volume ( cm³) Total volume Of HCl used ( cm³) Rough 0.2 30.3 30.1 1 6.3 35.8 29.5 2 2.7 32.0 29.3 Average 29.6 CONCLUSION Method 1 [IMAGE]2Li (s) + 2H20(l) LiOH(aq) + H2(g) Number of moles of Hydrogen. Volume of hydrogen gas was 185 cm³. Weight of Lithium was 0.11g. N = __V__ _185_ = 0.0077 MOLES 24000 24000 Number of moles of Lithium.
Reaction 2: H = 50 x 4.18 x -10.3" H = -2152.7 This value is for 1.37g of calcium oxide, not 56.1g, which is its relative molecular mass. Therefore: H =
reaction by seeing how much of it is used up in a particular amount of
And the symbol equation for it is:. Na2S2O3 + 2HCl, S + SO2 + 2Na + H2O. Before conducting my experiment, I will research into, amongst other things, the factors that affect the rate of a reaction. This is so that I may have enough information to understand the effect of temperature on the rate of a reaction and also gain appropriate understanding to make a suitable prediction as to what the outcome of my experiment will be. Reactions occur when the particles of reactants collide together continuously.
Figure 1: Simple batch homogenous reactor. [Fogler, H. S. (2010, November 22). Essentials of Chemical Reaction Engineering: Mole Balances. Retrieved April 24, 2014, from Pearson Education: http://www.informit.com/articles/article.aspx?p=1652026&seqNum=3]
In this experiment three different equations were used and they are the Stoichiometry of Titration Reaction, Converting mL to L, and Calculating the Molarity of NaOH and HCl (Lab Guide pg. 142 and 143).
Limiting reactant means in chemistry we have a certain amount of one elements combined with two amounts of another element to product x certain amount of product. But because one is more than the other we will run out before we use up the same equal amounts of each elements.
= ((M4g) - (M4(( S - ut ) / ( 0.5 t^2 )) - (FFr)) (R2+R1) / (( S - ut ) / ( 0.5 t^2 )) / R2
Therefore, the equation shows the same as the graphics, that the mass remaining decreases exponentially with time.