Question

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

Answer 1

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 convert O₂ moles into Cs₂O moles, using the stoichiometric coefficients of the balanced reaction:

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

Now we convert 2.125 moles of Cs₂O into grams, using its molar mass:

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

598.85 g is the theoretical yield. Finally we proceed to calculate the percent yield:

• 199 / 598.85 * 100% = 33.23 %