Limiting Reactant/Percent Yield Practice Problems Rule: If amounts of both (or a

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Limiting Reactant/Percent Yield Practice Problems Rule: If amounts of both (or all reactants are given you must first determine the limiting reactant before calculating theoretical yield and % yield 1. Consider the reaction 1:04g)+5 CO(g) 5 CO(g)l2g a) If 80.0 grams of 1:Os reacts with 28 0 grams of CO, determine the mass of l formed b) if only 30 0 g of l2 was produced. determine the peroent yiekd 2. 2n and S react to form Znc (l) Sufide, Zns al Determine the imiting reactant 500 g of Zinc reacted with 60 0 g of Sulfur b) Haw many grams of Zn5 will be farmed? c) How many grams of excess reactant wil remain after the reacion is over? d) calculate the percent yield if your actual yield was 38 0 g of Zns 3. Aluminum sulfte reacts with sodium hydraxide to form socium suifite and aluminum hydroxide accordng to the following equation a) Determine the linting reactant if 200 g of AISOh reacts with 2?0g of NaOH b) Determine the number of moles and mass of AVOHh produced c) Determine the number of moies and mass of Na S0, produced d) Determne the grams of excess reagent left over in the reaction 4. Siver 11) Nitrane reacts with Iron Ill, chioride to form Stver (1) chlarde and Iron (1) ntrate. 1 250 g of siver Il) nitrate reacted with 45 0 g of iran I1) chloride determine the limiting reactant and the grams for Siver (ll chionde formed in the resaction. If the acual yield was 20.0g calclale the percent yeld 5 f 150.0g of Magresum Hydroxide reacted wih 135 5 g of Hydrochionic Acid detoarmine the Imnng reactant for the reacion if the amount of Magnesium Chionide fomed wars 1200 g, calculabe the Percent Yeld 6 Consider the reaction CaH Br, cacuiale the percene yald of the reaction Consider the reaction of CO, + Br2 ?C'HsBr + HBr What is the theoretical yield of Coh 8r 84 0g of CeHe react with 146.0g of B? 7 If the actual yield of GoHsBr is 127.39, what is the peroent yeld?

Explanation / Answer

1) a) 1 mole of I2O5 (333.81 g) requires 5 moles (5*28 g) CO
80 g I2O5 requires (80 x 5*28 g)/ 333.81 = 33.55 g CO
but we have only 28 g CO, here CO is limiting reactant and it will decide the amount of the I2 formed
   5*28 g CO forms 1 mole of I2 (253.8g)
28 g CO forms (28 g x 253.8 g) / (5*28 g) = 50.76 g I2
b) From the above , theoritical Yiled of I2 = 50.76 g
  but actual yield = 30 g
   % Yield = (30g / 50.76 g ) x 100 = 59.1 %

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