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adelina 88 [10]
3 years ago
6

34 g of O2 are reacted with excess Cs, causing a production of 199 g of Cs2O. What is the percent yield of this

Chemistry
1 answer:
PtichkaEL [24]3 years ago
3 0

Answer:

33.23 %

Explanation:

  • 4 Cs + O₂ → 2Cs₂O

First we convert 34 g of O₂ into moles, using its molar mass:

  • 34 g O₂ ÷ 32 g/mol = 1.0625 mol O₂

Then we <u>convert O₂ moles into Cs₂O moles</u>, using <em>the stoichiometric coefficients of the balanced reaction</em>:

  • 1.0625 mol O₂ * \frac{2molCs_2O}{1molO_2} = 2.125 mol Cs₂O

Now we <u>convert 2.125 moles of Cs₂O into grams</u>, using its <em>molar mass</em>:

  • 2.125 mol Cs₂O * 281.81 g/mol = 598.85 g Cs₂O

598.85 g is the theoretical yield. Finally we proceed to <em>calculate the percent yield</em>:

  • 199 / 598.85 * 100% = 33.23 %
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