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Instructions on how to calculate the limiting reactant, theoretical yield, actual yield, and percent yield in a chemical reaction using the example of sodium iodide (nai) and bromine (br2) reaction. It also includes a step-by-step calculation for identifying the limiting reactant when sodium (na) reacts with ferric chloride (fecl3).
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VOCABULARY
IDENTIFY THE LIMITING REACTANT WHEN 8.7 g Na REACTS WITH 9.3 g FeCl 3 ? 3 Na + 1 FeBr 3 3 NaBr + Fe 8.7 grams Na x 1 mole Na x 3 moles NaBr x 102.89 grams NaBr = 38.9g NaBr 22.990 g Na 3 mole Na 1 mole NaBr 9.3 grams FeCl 3 x 1 mole Na x 3 moles NaBr x 102.89 grams NaBr =17.7 g NaBr 161.83 g FeCl 3 1 mole FeCl 3 1 mole NaBr FeCl 3 is the limiting reactant because it makes a smaller amount of product (NaBr). 17.7 grams NaBr is less then 38.9 g NaBr