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, N24(l), and disitroget tetroxide, N2O1(l), which ignite to form nitrogen gas and water vapor: 2 N2H(l) + N2O(l) rightarrow 3N2(g) + 4H2O(g)In a certain experiment 100 grams of N2H2(l)is allowed to reset with 200 grams of N2O(l) N = 14.0 g/mol, O = 16.0 g/mol, H = 1.00 g/mol follow steps alpha through f below and fill your answers in the table: Find the molar masses (g/mol) for the reactants and products, and fill your answers in the table. Find the initial number of moles for each of the reactants and products g x mol/theta = moles Fill your answers in the table Use the initial modes 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 result you obtained from the excess reactant since those values are impossible to produce. Fill your answers in the table. After the reaction in completed, none of the limiting reactant will remain. The limiting reactant was completely consumed in the reaction. Calculate the model and grams of the excess reactant that remain. Fill you answers in the table. Final mo excess reactant initial mol excess reactant-g/moles product Show that the law of conservation of mass is obeyed by your results. masses of reactants = masses of products + mass of remaining excess reactants

Explanation / Answer

Final mass : to find this we use final moles and molar mass

N2H4 : 0 mol = 0 gl

N2O4 = 0.61 x 92g/mol = 56.12 g

N2 = 4.68 x 28g/mol = 131.04 g

H2O = 6.25 x 18 g/mol = 112.5 g

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e)

Final number of moles of N2O4:

Number of moles of N2O4 present initially = 2.17 mol

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

Final mass = 0.61 x 92 = 56.12

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f)

Sum of masses of products = mass of N2 + mass of H2O + mass of excess N2O4

= 131.04 g + 112.5 g + 56.12 = 299.66

sum of masses of reactants = 100 + 200 = 300 g

So, the sum of masses of reactants = sum of masses of products + mass of remaining excess reactants.

Law of conservation of mass is proved.

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