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