com/myctieviewaig ssigementProb-963 Exercise 8.68 15 of 20 ?PartA Calculabe the

Question

com/myctieviewaig ssigementProb-963 Exercise 8.68 15 of 20 ?PartA Calculabe the mass of N H, that will be in the reaction vessel once the reactants have reacted as much as posstn Aasume 100% yed Consider the reaction between NHs and NO Express your answer with the appropriate wnits. A reaction vessel intalily contains 30.0g NaH, and 74 9 g of N,0s Hint: The limiting reactant i completely consumed, but the reactant in ecessis not. Use the amount of imiting reactant to determine the amount of products that form and the amount of the Value Units n excess hat remains ater complete Part B Calculate he mass of N2O·that wd be in the readan vessel orce the reactards have reached much as possble Assune 100% yield Express your answer with the appropriate units ??012018

Explanation / Answer

Mass of N2O4 = 74.9 g.

Molar mass of N2O4 = 2 ( 14 ) + 4 ( 16 ) = 92.0 g/mol

Moles of N2O4 = mass / molar mass = 74.9 / 92.0 = 0.814 mol

Mass of N2H4 = 30.0 g.

Molar mass of N2H4 = 32.0 g/mol

Moles of N2H4 = 30.0 / 32.0 = 0.938 mol

From the balanced equation,

2 mol of N2H4 needs 1 mol of N2O4

Then, 0.938 mol of N2H4 needs 1 * 0.938 / 2 = 0.469 mol of N2O4 ( < .0814 mol )

SO, N2H4 is limiting reagent

And N2O4 is excess reagent.

(A)

No N2H4 is remained as it is limiting reagent undergoes 100 % consumption.

(B)

Moles of N2O4 remained = 0.814 - 0.469 = 0.345 mol of N2O4

Mass of N2O4 remained after reaction = moles * molar mass = 0.345 * 92.0 = 31.8 g.

(C)

From the balanced equation,

2 mol of N2H4 forms 3 mol of N2

Then, 0.938 mol of N2H4 forms 3 * 0.938 / 2 = 1.41 mol of N2

Therefore,

Theoretical yield of N2 = moles * molar mass = 1.41 * 28.0 = 39.4 g.

(D)

From the balanced equation,

2 mol of N2H4 forms 4 mol of H2O

then, 0.938 mol of N2H4 forms 0.938 * 4 / 2 = 1.88 mol of H2O

Therefore,

Theoretical yield of H2O = moles * molar mass = 1.88 * 18.0 = 33.8 g. of H2O

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