Calculating Percent Yield and Identifying Limiting Reactants in Stoichiometry, Schemes and Mind Maps of Stoichiometry

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).

Typology: Schemes and Mind Maps

2021/2022

Uploaded on 09/12/2022

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STOICHIOMETRY II
CALCULATING LIMITING REACTANT, EXCESS REACTANT, THEORETICAL YIELD,
AND PERCENT YIELD
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STOICHIOMETRY II

CALCULATING LIMITING REACTANT, EXCESS REACTANT, THEORETICAL YIELD,

AND PERCENT YIELD

VOCABULARY

 LIMITING REACTANT – the reactant that runs out first. The limiting reactant controls

how much product that can be made because when the reactant runs out the

reaction stops.

 EXCESS REACTANT-the reactant there is plenty of. The reactant there is leftovers of

because all of the excess reactant isn’t used.

 THEORETICAL YIELD – the maximum amount of product that will be made from a

chemical reaction.

The theoretical yield is calculated from a grams to grams (3 step) calculation.

The theoretical yield is dependent on the limiting reactant.

 ACTUAL YIELD – the mass of product made experimentally in the lab setting.

 % YIELD – is a mathematical comparison of how much product was made

compared to how much should have been made.

% yield = ACTUAL YIELD (100)

THEORETIAL YIELD

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