Question

Given the following chemical equation, if 162.0 grams of potassium reacts with excess oxygen gas and was found to produce 172 grams of potassium oxide, what is the percent yield for this reaction?
4 K(s) + O2(g) arrow 2K2O(s)

Answer 1

hmm

Answer 2

i honestly dont know this one

Answer 3

\[Number of Moles=Mass(g)/Molar Mass(g/mol) \]Atomic Weight of K(potassium)=39 (g/mol)
Given Mass of K(potassium):162 gram
\[N of K(potassium)=162/39=4.15\]
\[4 K(s) + O2(g) \rightarrow 2K2O(s) \]
Look for the mole ratio of K to K2O from the balanced chemical equation. 4:2 is the mole ratio
\[4:2\]
\[4.15:x mole of K2O\]
\[x=(4.15*2)/4=2.076 \]
Use again Number of Moles=Mass(g)/Molar Mas(g/mol) but rearrange the equation to
\[Mass(g)=Molar Mass(g/mol)*Number of Moles \]
Atomic Weight of O and K
O:16 gram K:39 gram
\[ Molar Mass of K2O: (39*2)+(16*1)=94 gram \]
\[N of K2O=2.076*94=195.44 g\]
(Actual Yield/Theortical Yield)*100=Percentage Yield
(162/195.4)*100=82.9%

Answer 4

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