This question concerns stoichiometry and the limiting reactant.

Question

On one of the Apollo lunar missions, N2O4 powered the lunar landing module as it ascended from the surface of the moon. During the ascent, 1.200 times 10^3 kg N2H2, 1.000 times 10^3 kg (CH3)2N2H2, and 4.500 times 10^3 kg N2O2 were available in the following reaction:

2 N2H4 + (CH3)2N2H2 + 3N2O4 to 6N2 + 2CO2 + 8H2O

Determine the following:

A) Which of the compounds was consumed first in this reaction?

B) How many kilograms of wastewater were spewed into space by the reaction?

C) How many nitrogen atoms were left in space by the reaction?

Answer 1

2 N₂H₄ + (CH₃)₂N₂H₂ + 3N₂O₄ ==> 6N₂ + 2CO₂ + 8H₂O

1.2x10³kg...1.0x10³kg.....4.5x10³kg....

First thing to do is find which reactant is LIMITING

An easy way to do this is to divide moles of each reactant by the corresponding coefficient in the balanced equation. Whichever value comes out less represents the limiting reactant. Thus...

For N2H4: 1.200 x10³kg x 1000 g / kg x 1 mol / 32.05 g = 37441 mols (÷2->18721)

For (CH₃)₂N₂H₂: 1.00x10³kg x 1000 g / kg x 1 mol / 60.10 g = 16639 mols (÷1->16639)

For N₂O₄: 4.50x10³ kg x 1000 g / kg x 1 mol / 92.01 g = 45647 mols (÷3->15216)

(A). Since 15216 is the lowest value, this tells us that N₂O₄ is the limiting reactant and moles of N₂O₄ (45647) will determine the amount of products formed and the amount of reactants remaining after the reaction. So answer to (A) is that N₂O₄ will be used up first.

(B). Kg of H₂O formed and spewed into space? Use the stoichiometry of the balanced equation:

45647 mols N₂O₄ x 8 mol H₂O/3 mol N₂O₄ = 121725 mols H₂O x 18.02 g/mol x 1 Kg/1000 g = 2191 kg

(C). Nitrogen atoms (N) left in space:

45647 mol N₂O₄ x 6 mol N₂ / 3 mol N₂O₄ = 91294 mol N₂

91294 mol N₂ x 6.02x10²³ molecules N2 / mol x 2 N atoms / molecule = 1.099x10²⁹ N atoms

(be sure to check all of the math)

Also, you should do the calculations and round to the appropriate number of sig.figs. required by your instructor.

Answer 2

This problem concerns mole ratios and limiting reactants.

The balanced equation gives you mole ratios of the substances involved in the reaction, just like a cake recipe gives you the ratios of ingredients you must use to make the cake(s).

Part A of this questions asks for the limiting reactant, which is the reactant that gets completely used up in the reaction If you need 1 cup of flour and 2 eggs to make a cake, and you have 3 cups of flour, but only 4 eggs, then eggs are limiting the number of cakes you can make. You will use all your eggs to make 2 cakes, and have some flour left over. To answer part A, you need to first convert the given masses into moles, and then use you mole ratios to determine which reactant is limiting.

Part B requires that you use mole ratios to predict the amounts of product that will be produced by the reaction. You should have already converted your reactant masses into moles in step A of this problem; you should have already determined the limiting reactant. How many moles of the limiting reactant will be used by the reaction? Use the mole ratio of the limiting reactant to the product in question (water, in this case) to determine how many moles of water will be produced. Finally, convert moles of water in to kgs using your molar mass

Part C requires that you do the same thing you did for water in part B, except now do it for Nitrogen, and instead of converting from moles of nitrogen to mass, convert to atoms using avogadro's number

Let me know if you have any questions!