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Semmy [17]
3 years ago
9

When 150. g zinc sulfide are burned in excess oxygen, 68.5 g of zinc oxide are actually produced, along with sulfur dioxide. Det

ermine the percent yield of zinc oxide after finding the theoretical production of zinc oxide.
**Show the balanced equation before you begin.**
Chemistry
1 answer:
Bumek [7]3 years ago
4 0

%yield = 54.6%

<h3>Further explanation</h3>

Percent yield is the compare of the amount of product obtained from a reaction with the amount you calculated

(theoretical)

General formula:

Percent yield = (Actual yield / theoretical yield )x 100%

<h3 />

Reaction

2ZnS+3O₂ ⇒ 2ZnO+2SO₂

MW ZnS = 97.474 g/mol

  • mol ZnS

\tt \dfrac{150}{97.474}=1.54

MW ZnO = 81.38 g/mol

  • mol ZnO (from mol ZnS as limiting reactant, O₂ excess)

\tt \dfrac{2}{2}\times 1.54=1.54

  • Actual ZnO produced

\tt 1.54\times 81.38=125.33~g

Theoretical production = 125.388

  • %yield

\tt \dfrac{68.5}{125.33}\times 100\%=\boxed{\bold{54.6\%}}

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Suppose the half-life is 9.0 s for a first order reaction and the reactant concentration is 0.0741 M 50.7 s after the reaction s
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<u>Explanation:</u>

The equation used to calculate half life for first order kinetics:

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k=\frac{0.693}{9}=0.077s^{-1}

Rate law expression for first order kinetics is given by the equation:

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k = rate constant  = 0.077s^{-1}

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Putting values in equation 1, we get:

0.077=\frac{2.303}{50.7}\log\frac{[A_o]}{0.0741}

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