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

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## Limiting Reagents

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

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