In Problem 6.40, hydrazine reacted with oxygen according to the f...

Question

In Problem 6.40, hydrazine reacted with oxygen according to the following (unbalanced) equation:

N2H4(l) + O2(g) → NO2(g) + H2O(g)

If 75.0 kg of hydrazine are reacted with 75.0 kg of oxygen, which is the limiting reagent?

Accepted Answer

Welcome back everyone. The high energy fuel, 11 dimethyl hydrazine reacts spontaneously with oxygen to give gaseous products. According to the following reaction, where 11 dimethyl hydrogen reacts with oxygen gas to form carbon dioxide, gas, water vapor and nitrogen gas as products. If 50 kg of this fuel is reacted with the same mass of oxygen, which of the two reactants will serve as the limiting reagent because the given equation is not balanced. We're going to need to start by comparing the number of atoms on the reactant versus product side. Starting out by listing out the atoms involved. We have nitrogen, hydrogen, carbon and oxygen on both sides of the reaction. And listing out the number of atoms beginning with nitrogen. On the reactant side, we have a subscript of two. So we have two atoms of nitrogen. Then we have hydrogen which has within the chemical formula of 11 dimethyl hydrazine, a subscript of two along with a subscript of three within parentheses for the second hydrogen with another subscript of two outside of the parentheses. So we would have three times two, which would give us six atoms of hydrogen plus the additional two, which would give us a total of eight atoms of hydrogen on the reactant side. Then we have carbon on the reactant side where we would observe by the subscript of two on the outside of the parentheses, a total of two carbon atoms. And then for oxygen on the reactant side, we also have two atoms. Now moving on to listing out the atoms on the product side, we have two atoms of nitrogen. Then for hydrogen, on the product side, we have two atoms of hydrogen. And then for carbon on the product side, we have one atom of carbon and then a total of three atoms of oxygen. On the product side, we're going to start by balancing out non oxygen and non hydrogen atoms first. So starting out with the imbalanced with the imbalance between the carbon atoms because we have two on the reactant side and one on the product side, let's start by placing a coefficient of two in front of our products of carbon dioxide gas. That's now going to give two atoms of carbon on both sides of the reaction and will change the number of oxygen on the product side to a total of five atoms. Now, in order to then balance out, we'll start with hydrogen because we only have two atoms of hydrogen on the product side. And eight atoms of hydrogen on the reactant side, we can then place a coefficient of four in front of our product water, we'll use a different color. So four in front of water vapor on the product side is going to give us now four moles of hydrogen rather not four. But when we multiply by the subscript of two in water, on the product side, that would give us actually eight atoms of hydrogen on the product side. And that will balance out with the eight atoms of hydrogen already present on the reactant side. However, now that changes our total number of oxygen on the product side to a total of eight atoms of oxygen. And so to mend that we would then place a coefficient of four in front of the reactant oxygen gas, which will now give a total of when we multiply by the subscript of 28 atoms of oxygen on both sides of the reaction. And now we have a fully balanced equation. So given the mass of our reactant of kg of 11 dimethyl hydrazine, we're going to start by utilizing a conversion in dimensional analysis will where we will multiply to go from kilograms of 11 dimethyl hydrazine in the denominator to then grams of 11 dimethyl hydrazine in the numerator. We're going to recall that our prefix kilo tells us that we have an equivalent of 10 to the third power of our base unit gram canceling out kilograms. We have grams of this reactant and we're going to figure out how much based on our mass of this reactant, how much product is produced? And we can choose between any of the three products in our reaction. So let's multiply then to go next from grams of 11 dimethyl hydrazine in the denominator to then moles of this reactant in the numerator. This is where we're going to utilize our periodic tables. And we're going to calculate the molar mass of this reactant. 11 dimethyl hydrazine using our tables in our textbooks, we would find a molar mass of 60.1 oh four g of 11 dimethyl hydrazine per mole in the numerator, canceling out grams of 11 dimethyl hydrazine. We are now at moles and we're going to refer to the ratio between 11 dimethyl hydrogen and one of our products. Let's go with carbon dioxide. So we would observe that based on our balanced equation, we have for every one mole of 11 dimethyl hydrogen reacted, we have two moles of carbon dioxide produced and that is our molar ratio. So we will multiply by that. Next, we will multiply by that. Next. Where in our denominator, we will go from moles of 11 dimethyl hydrazine to then moles of the product CO2 and plugging in that molar ratio using the stric coefficients. We said we have one, a 1 to 2 molar ratio. So canceling out now, moles of 11 dimethyl hydrazine, we now have moles of the product produced and we'll just stop here, simplifying in our calculators, we would find the moles of product CO2 produced being 1663. moles of carbon dioxide produced. And this is produced from the first reactant, 11 dimethyl hydrazine. Now, we're going to compare to moles of carbon dioxide produced by the second reactant, which is oxygen gas. We're given that the oxygen gas is reacted as the same mass as 11 dimethyl hydrogen. So we would have also 50 kg of oxygen in which we will multiply to convert from kilograms of oxygen in the denominator to grams of oxygen in the numerator where just as before, for one kg, we have an equivalent of 10th of the third power grams in the numerator. So canceling out kilograms of oxy, we are at grams of oxygen and we will next multiply to incorporate the molar mass of oxygen where we would have grams of oxygen in the denominator per mole of oxygen in the numerator. And utilizing our periodic tables in our textbooks, we would reference the molar mass of sorry, a 32 0.0 g of oxygen per mole of oxygen and recall that oxygen is diatomic. That's why it has that subscript of two. So we have grams per mole of oxygen and canceling out grams of oxygen. We will then focus on multiplying next to go from a molar ratio where moles of oxygen is in the denominator and moles of our product being made CO is in the numerator. And this is where we go back to the coefficient and refer to our balanced equation where for every four moles of oxygen reacted, we have two moles of carbon dioxide produced. So in our denominator, we have four moles of oxygen. And then we have two moles of CO2 in the numerator, canceling out moles of oxygen. We are left with moles of our product CO2 and simplifying in our calculators, we would find the moles of carbon dioxide produced by oxygen is 781.25 moles of produced, sorry moles of CO2 produced and based on the two reactants and their corresponding amounts of carbon dioxide product produced because oxygen makes a smaller amount of product as 781.25 moles. The reactant that gives the smaller number of moles is going to be our limiting reactant. So we'll say that therefore, oxygen is the limiting reactant because it produces again the smaller amount of product. And so our final answer is going to be that 02 or oxygen is the limiting reactant. I hope that this made sense and let us know if you have any questions.

In Problem 6.40, hydrazine reacted with oxygen according to the following (unbalanced) equation: N2H4(l) + O2(g) → NO2(g) + H2O(g) If 75.0 kg of hydrazine are reacted with 75.0 kg of oxygen, which is the limiting reagent?

Key Concepts

Limiting Reagent

Stoichiometry

Molar Mass

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