A fuel mixture used in the early days of rocketry is composed of two liquids hyd

Question

A fuel mixture used in the early days of rocketry is composed of two liquids hydrazine, N2H (I). and dinitrogen tetroxide. N2O4 (l).which ignite to form nitrogen gas and water vapor: In a certain experiment 100 grams of N2H4(l) is allowed to react with 200 grams of N2O(l Follow steps a through/ below and fill your answers in the table Find the molar masses in mo!) for the reactants and products, and fill your answers in the table. Find the initial number of moles for c ach of the reactants and products g times moles/g = modes Fill your answers in the table. Use the initial moles of each reactant and the reaction coefficients to determine how many moles of products can be theoretically formed according to each reactant. One of the reactants is the "limiting reactant" because it yields fewer moles of products than the other reactant. Use the limiting reactant results to select the moles that can form for each product "Discard" the results you obtained from the excess reactant, since those values arc impossible to produce. Fill your answers in the table

Explanation / Answer

Molar mass :

N2H4 : 2(14) + 4(1) =32 g/mol

N2O4 = 2(14) + 4(16) = 92g/mol

N2 = 2(14) = 28g/mol

H2O = 2(1) + 1(16) = 18 g/mol

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INitial mass

N2H4 : 100 g

N2O4 = 200 g

N2 = 0

H2O = 0

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Initial number of moles

N2H4 : 100g / 32 g/mol = 3.125 mol

N2O4 = 200 / 92g/mol = 2.17 mol

N2 = 0 / 28g/mol = 0

H2O = 0 / 18 g/mol = 0

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Final number of moles

First, we need to find the limiting reagent. the moles of the reactants are divided by the stoichiometric coefficients present in the chemical reaction.

N2H4 = 3.125 mol / 2 = 1.56 mol

N2O4 = 2.17mol / 1 = 2.17 mol

So, as the moles of N2H4 are less, N2H4 is the limiting reagent.

So, the final number of moles of N2H4 = 0 mol

Final number of moles of N2O4:

2 mol of N2H4 react with 1 mol N2O4. ( according to the equation)

So, 3.125 mol of N2H4 react with (3.125 / 2) = 1.56 mol of N2O4.

So, left out N2O4 = 2.17 - 1.56 = 0.61 mol

So, final number of moles of N2O4 = 0.61 mol

from the chemical equation, 2 mol N2H4 give 3 mol N2.

So, 3.125 mol N2H4 give (3.125 x 3) /2 = 4.68 mol of N2.

Final number of moles of N2 = 4.68

from the chemical equation, 2 mol of N2H4 give 4 mol H2O.

So, 3.125 mol N2H4 give (3.125 x 4) / 2 = 6.25 mol H2O.

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