Stoichiometry: Mole Ratios, Limiting Reactants, and Percent Yield, Summaries of Stoichiometry

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STOICHIOMETRY

HONORS CHEMISTRY

MOLE RATIO

A mole ratio is the ratio of coefficients used to compare amounts of reactants and products.

1 ZnCl 2 (aq) + 2 NaOH (aq) 1 Zn(OH) 2 (aq) + 2 NaCl (aq)

What is the mole ratio of NaOH to NaCl? 2:2 or 2 moles NaOH: 2 moles NaCl

What is the mole ratio of ZnCl 2 to NaCl? 1:2 or 1 mole ZnCl 2

1 mole NaCl

note that the coefficient for ZnCl 2 is listed first, and the

coefficient for NaCl is listed second because that is the

order of the substances in the question.

GRAM TO GRAM 3 step conversion 4 FeS + 7 O 2 → 2 Fe 2 O 3 + 4 SO QUESTION 2: If 15.88 grams O 2 of oxygen react, what mass of SO will be made? molar mass SO = 32 + 16

15.88 grams O 2 x 1 mole O 2 x 4 moles SO x 48 grams SO = (15.88 grams O 2 )(1 mole O 2 )(4 moles SO)(48 g SO) = 13.20 g SO

32.998 g O 2 7 moles O 2 1 mole SO (32.998 g O 2 )(7 moles O 2 )(1 mole SO)

Molar mass O 2 = 15.999x2 15.88 x 1 x 4 x 48 ÷ 32.998 ÷ 7 ÷ 1 = 13.20 g SO Observe the pattern of units: Grams A x 1 moles A x moles B x grams B = grams B molar mass A moles A molar mass B

GRAM TO MOLE 2 step conversion

4 FeS + 7 O 2 → 2 Fe 2 O 3 + 4 SO

QUESTION 3: If 15.88 grams O 2 of oxygen react, what amount in moles of SO will be made?

15.88 grams O 2 x 1 mole O 2 x 4 moles SO = (15.88 grams O 2 )(1 mole O 2 )(4 moles SO) =

32.998 g O 2 7 moles O 2 (32.998 g O 2 )(7 moles O 2 )

15.88 X 1X 4 ÷32.998 ÷ 7 = 0.27 moles SO

The gram to mole 2 step conversion is the same as the 3 step conversion, but excludes the third

step, leaving the value in moles.

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

Calculate the percentage yield of NaBr if 20 grams NaI and 12 g Br 2 react and form 12.3 g NaBr in the lab? 2 NaI + Br 2 2 NaBr + I 2 Start with 3 step converting grams of reactant to grams of product. 20 grams NaI x 1 mole NaI x 2 moles NaBr x 102.89 grams NaBr = 13.7 g NaBr 149.89 g NaI 2 mole NaI 1 mole NaBr limiting reactant molar mass NaI = 22.990+126.904 molar mass NaBr = 22.990+79. 12 grams Br 2 x 1 mole Br 2 I x 2 moles NaBr x 102.89 grams NaBr = 15.45 g NaBr 159.81 g Br 2 1 mole Br 2 1 mole NaBr excess reactant molar mass Br 2 = 79.904 x 2 molar mass NaBr = 22.990+79. Identify the smaller value as theoretical yield. 13.7 g NaBr is the smaller value, therefore the theoretical yield. 15.45 grams of NaBr will never be made because once 13.7 g NaBr is made the reaction runs out of NaI, therefore the reaction stops producing NaBr product. NaI is known as the limiting reactant. Solve for percent yield. ACTUAL YIELD (100) = 12.3 grams NaBr (100) = 89.8 % THEORETICAL YIELD 13.7 grams NaBr Hint: actual yield will always be given in problem, theoretical yield will always be calculated with a 3 step grams to grams conversion