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Limiting Reagents

Limiting Reagents

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Limiting Reagents

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  1. Limiting Reagents We be cooking…. Oh crap I only have ½ cup of Sugar!

  2. Brownies again… Only have ½ cup of sugar…bummer….what to do? ½ cup butter 2 oz of chocolate 1 cp sugar 2 eggs 1 tsp vanilla 2/3 cp flour ½ tsp baking powder ¼ tsp salt 24 Brownies Need to do some Stoiching to figure out how much of other ingredients if only have ½ cup sugar…. Sugar is the limiting reagent Everything else in excess.

  3. Limiting Reagent • Demo • Solid zinc is added to 6.00 M HCl • Zinc will all dissolve but…. • pH is still below 7…meaning still have H+ ions present • Which reagent is limiting? • Zinc • Which reagent is excess? • HCl

  4. Stoichiometry with Limiting Reagent • How to determine the limiting and excess reagent • 1. Get both reagents into the unit moles • 2. Divide those moles found by the coefficient of the reagent • 3. The smaller number is the limiting reagent (LR) • 4. The larger number is in excess • 5. All stoich problems will be determined by the limiting reagent (LR)…therefore the given for DA will be the LR

  5. Example with Limiting Reagent 7.24 moles of magnesium is added to 3.86 moles of oxygen gas to make MgO. Which reactant is used up? Excess? 2Mg(s) + O2(g)  2MgO(s) 7.24 mol Mg = 3.62 mol of Mg Mg is LR 2 3.86 mol O2 O2 is excess = 3.86 mol of O2 1 7.24 mol of Mg controls the reaction, therefore 7.24 mol will be your given for any DA

  6. Example with Limiting Reagent 7.24 moles of magnesium is added to 3.86 moles of oxygen gas to make MgO. How many grams of MgO is produced? How many grams of oxygen is needed? 2Mg(s) + O2(g)  2MgO(s) 7.24 mol Mg 2 mol MgO 40.304 g MgO = 292 g of MgO 2 mol Mg 1 mol MgO 7.24 mol Mg 1 mol O2 31.998 g O2 = 116 g of O2 2 mol Mg 1 mol O2

  7. Remember… • Must work with the Limiting Reagent in Stoich • Must get both reagents into moles • Divide each by their coefficient • The smaller number is the LR…controls the reaction • This will be your given in DA

  8. Problem If 7.56 gram of iron metal are placed in 100. mL of a 1.00 M hydrochloric acid, HCl, solution, hydrogen gas and iron(II) chloride are produced. Which reactant is limiting? How many grams are in excess? How many grams of each product is formed? • Balance chemical reaction Fe(s) + 2HCl(aq)  H2(g) + FeCl2(aq)

  9. Problem If 7.56 gram of iron metal are placed in 100. mL of a 1.00 M hydrochloric acid, HCl, solution, hydrogen gas and iron(II) chloride are produced. Which reactant is limiting? How many grams are in excess? How many grams of each product is formed? Fe(s) + 2HCl(aq)  H2(g) + FeCl2(aq) 2. Find moles of each reactant to find LR 7.56 g Fe 1 mol Fe = 0.135 mol Fe = 0.135 mol Fe 55.845 g 1 LR 100. mL 1 L 1.00 mol HCl = 0.100 mol HCl 1000 mL 1 L 2 = 0.0500 mol HCl

  10. Problem If 7.56 gram of iron metal are placed in 100. mL of a 1.00 M hydrochloric acid, HCl, solution, hydrogen gas and iron(II) chloride are produced. Which reactant is limiting? How many grams are in excess? How many grams of each product is formed? Fe(s) + 2HCl(aq)  H2(g) + FeCl2(aq) 3. Find grams of excess (find grams you actually need) 0.100 mol HCl is LR 0.100 mol HCl 1 mol Fe 55.845 g Fe = 2.79 g of Fe 2 mol HCl 1 mol Fe 7.56 gram Fe – 2.79 gram Fe = 4.80 gram Fe excess

  11. Problem If 7.56 gram of iron metal are placed in 100. mL of a 1.00 M hydrochloric acid, HCl, solution, hydrogen gas and iron(II) chloride are produced. Which reactant is limiting? How many grams are in excess? How many grams of each product is formed? Fe(s) + 2HCl(aq)  H2(g) + FeCl2(aq) 3. Find grams of each product (H2 and FeCl2) 0.100 mol HCl is LR 0.100 mol HCl 1 mol FeCl2 126.745g FeCl2 = 6.34 g of FeCl2 2 mol HCl 1 mol FeCl2 0.100 mol HCl 1 mol H2 2.0158 g H2 = 0.101 g of H2 2 mol HCl 1 mol H2

  12. Review Steps for Stoiching… • Write and balancing chemical reaction • Find limiting reagent • Determine moles of each reactant • Divide the moles be their coefficient • The smaller number is LR • Use the moles of the LR to convert to anything else in the problem • Moleland!! • Convert to moles of another compound/atom • Convert to final unit (moles/grams/volume)

  13. Percent Yield Cu(s) + 2AgNO3(aq)  2Ag(s) + Cu(NO3)2(aq) How many grams of copper will be required to completely replace silver from 208 mL of 0.100 M solution of AgNO3? Molar mass of Cu Coefficients 208 mL 1 L 0.100 mol AgNO3 1 mol Cu 63.546 g Cu 1000 mL 1 L 2 mol AgNO3 1 mol Cu = 0.661 g of Cu

  14. Percent Yield 100% is maximum Percent yield measures how well you did the lab Compares lab measurement (actual) over theoretical amount (found using Stoich) % Yield = Actual amount (grams or moles) [Lab] Actual amount (grams or moles) x 100 Theoretical amount (grams or moles) [Stoich]

  15. Percent Yield If 12.5 grams of copper are reacted with an excess of chlorine gas, then 25.4 grams of copper(II) chloride, CuCl2(s), are obtained. Calculate theoretical amount and percent yield. Cu(s)+ Cl2(g)  CuCl2(s) Find grams CuCl2 Molar mass of CuCl2 Molar mass of Cu Coefficients 12.5 g Cu 1 mol 1 mol CuCl2 134.446 g CuCl2 63.546 g 1 mol Cu 1 mol CuCl2 = 26.4 g of CuCl2 25.4 g % Yield = x 100 = 96.4% 26.4 g