If you react 55g of hydrazine and 55g of oxygen gas, what is the maximum moles of nitrogen dioxide gas that can be made and what is the limiting reagent?
Write the equation and balance it.
I solve limiting reagent problems by working a simple stoichiometry problem twice; i.e., once with one reagent and the second with the other reagent. Each time you want to calculate the amount of product formed. You will get different answers for the product; the correct answer in limiting reagent problems is ALWAYS the smaller one and the reagent producing that value of the product is the limiting reagent.
To determine the maximum moles of nitrogen dioxide (NO2) gas that can be produced and the limiting reagent, we need to follow a few steps:
Step 1: Write and balance the chemical equation.
The balanced chemical equation for the reaction between hydrazine (N2H4) and oxygen gas (O2) to produce nitrogen dioxide gas (NO2) is:
2N2H4 + O2 → 4NO2 + 2H2O
Step 2: Find the molar masses of the reactants.
The molar mass of hydrazine (N2H4) = 32.04 g/mol
The molar mass of oxygen gas (O2) = 32.00 g/mol
Step 3: Calculate the number of moles for each reactant.
Moles of hydrazine (N2H4) = mass / molar mass = 55 g / 32.04 g/mol = 1.717 mol
Moles of oxygen gas (O2) = mass / molar mass = 55 g / 32.00 g/mol = 1.719 mol
Step 4: Determine the limiting reagent.
The limiting reagent is the reactant that will be completely consumed first and thus limits the amount of product formed. To find the limiting reagent, we compare the stoichiometric ratio between the reactants and the products.
From the balanced equation, we can see that the ratio of N2H4 to NO2 is 2:4, while the ratio of O2 to NO2 is 1:4. This means that for every 2 moles of N2H4, we will produce 4 moles of NO2, and for every 1 mole of O2, we will produce 4 moles of NO2.
Comparing the moles of hydrazine (N2H4) and oxygen gas (O2), we see that the molar ratio of N2H4 to O2 is approximately 1:1.
Since the stoichiometric ratio of N2H4 to NO2 is 2:4 (1:2), and the stoichiometric ratio of O2 to NO2 is 1:4, we can conclude that the limiting reagent is the one that produces fewer moles of NO2.
From the calculations above, we have:
Moles of NO2 produced from N2H4 = 1.717 mol × (4 mol NO2 / 2 mol N2H4) = 3.434 mol NO2
Moles of NO2 produced from O2 = 1.719 mol × (4 mol NO2 / 1 mol O2) = 6.876 mol NO2
Since 3.434 mol is less than 6.876 mol, the hydrazine (N2H4) is the limiting reagent.
Step 5: Determine the maximum moles of nitrogen dioxide (NO2) gas produced.
Using the stoichiometry from the balanced equation, we can conclude that the maximum moles of NO2 gas produced will be 3.434 moles.
So, the maximum moles of nitrogen dioxide (NO2) gas that can be produced is 3.434 mol, and the limiting reagent is hydrazine (N2H4).