Answer to Question #165973 in General Chemistry for jesus perez

Question #165973

How many moles of NO₂ would be produced from 4.5 mol of O₂ in the reaction below assuming the reaction has a 57.0% yield?

2 NO (g) + O₂ (g) → 2 NO₂ (g)


1
Expert's answer
2021-02-24T03:37:00-0500

From the reaction,

"2 NO (g) + O\u2082 (g) \u2192 2 NO\u2082 (g)"

One mole of "O_2" produces two moles of "NO_2" theoretically .

So, theoretical yield"(Y_t)" of "4.5" moles of "O_2" will be "4.5\\times 2= 9" moles of "NO_2" .

Let "Y_a" Be the actual yield of "NO_2."

Percentage yield "= \\frac{Y_a}{Y_t} \\times 100"

Substituting the values,

"57= \\frac{Y_a}{9} \\times 100"

"\\implies Y_a= \\frac{57\\times 9}{100}=5.13"

So, "5.13" moles of "NO_2" will pe produced from "4.5" moles of "O_2."



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