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Stoichiometry Limiting Problem

Stoichiometry Limiting Problem. ( ICE BOX METHOD). Initial-Change-End Box Method. Here are the Rules from the other Page STEP1= SET UP the ICE Box STEP 2- Find the moles, This is where you have to problem solve. STEP 3- Find X, find the moles of everything

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Stoichiometry Limiting Problem

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  1. Stoichiometry Limiting Problem ( ICE BOX METHOD)

  2. Initial-Change-End Box Method • Here are the Rules from the other Page • STEP1= SET UP the ICE Box • STEP 2- Find the moles, This is where you have to problem solve. • STEP 3- Find X, find the moles of everything • STEP 4- Answer the questions, convert moles to mass • The question- • You have 20.0 g of elemental sulfur, S, and 160.0 grams of O2. What mass of SO2 can be formed? How much reactant is left over? • S(s) + O2(g) ==> SO2(g)

  3. STEP1= SET UP the ICE Box below the balanced reaction. (I-Initial C-Change E-End). Using the coefficients, (the #'s in front of each compound) to determine the relative change. If there is a 2 in front, use 2X, if there is a 3 use 3X. Also reactants decrease in their amounts (-X), products will increase (+X).

  4. Step 2- Moles are the only thing allowed in the ice box. Find the moles of each compound and insert that value in its “I” box. • Moles S • 20.0g S X 1 mole/ 32.1g • =0.623 moles S • Moles O2 • 160.0g O2 X 1 mole/32.0g • =5.000 moles O2

  5. Step 3- Find X, one of the reactants is limiting, which means it runs out. You end up with 2 possible scenarios for this reaction. • if S runs out ==> 0.623 mol -X =O ; X is therefore 0.623 mol • if O2 runs out ==> 5.000 mol -X=O ; X is therefore 5.000 mol • Which ever reactant gives you the lower value for X is the limiting reactant and this X value is applied as X in your ICE BOX.

  6. Solve the Problem • (If you applied the incorrect, larger X value, will get a negative amount at the end, so go back and change it.)Now solve for everything (add or subtract down each column).

  7. Step 4- Answer the questions, you have them moles of everything. Convert to grams. • What mass of SO2 can be formed? • 0.623mol SO2X 64.1g/1 mol • =39.9g SO2 • How much reactant is left over? • 4.377mol O2X 32.0g/ 1 mol • =140.1g SO2

  8. Level 2- The equation now has coefficients. • If you are provided 200.g of sodium and 250. grams of iron (III) oxide, which substance is the limiting reactant? How many grams of Iron are produced? How much reactant is in excess? • 6Na + Fe2O3 --> 3Na2O + 2Fe

  9. STEP1= SET UP the ICE Box below the balanced reaction. (I-Initial C-Change E-End). • Using the coefficients, (the #'s in front of each compound) to determine the relative change. If there is a 2 in front, use 2X, if there is a 3 use 3X. Also reactants decrease in their amounts (-X), products will increase (+X).

  10. Step 2- Moles are the only thing allowed in the ice box. Find the moles of each compound and insert that value in its I box. • Moles Na • 200.g Na X 1 mole/ 23.0g • =8.70 moles Na • Moles Fe2O3 • 250.g Fe2O3 X 1 mole/ 159.7g • =1.57 moles Fe2O3

  11. Step 3- Find X, one of the reactants is limiting, which means it runs out. You end up with 2 possible scenarios for this reaction. • if Na runs out ==> 8.70 mol -6X =O ; X is therefore 1.45 mol • if Fe2O3 runs out ==> 1.57mol -X=O ; X is therefore 1.57 mol

  12. Which ever reactant gives you the lower value for X is the limiting reactant and this X value is applied as X in your ICE BOX. • Na is therefore limiting. • (If you applied the incorrect, larger X value, will get a negative amount at the end, so go back and change it.) • Now solve for everything (add or subtract down each column).

  13. Step 4- Answer the questions, you have them moles of everything. Convert to grams. • How many grams of Iron are produced? • 2.90 mol Fe X 55.85g/1 mol • =162g Fe • How much reactant is in excess? • 0.12 mol Fe2O3 X 159.7g/1 mol • =19.2g Fe2O3

  14. Work Cited • http://kentchemistry.com/

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