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Hydrazine gas (N2H4) is used as a rocket fuel. When hydrazine gas reacts with oxygen gas it produces nitrogen gas and water vapor (gas). If 30.0 grams of hydrazine gas are reacted with 45.0 grams of oxygen gas, what is the theoretical yield of water vapor (gas) that can be produced?Equation: ?Limiting Reactant: ?Excess: ?Grams of target product produced: ?

User Widlyne
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Our unbalanced reaction is:

N₂H₄ + O₂ ----> N₂ + H₂O

Now we can balance it. We can change the coefficient for water and write a 2 in front it and the equation will be balanced. The balanced equation is:

N₂H₄ + O₂ ----> N₂ + 2 H₂O

Then we are told that 30.0 g of N₂H₄ will react with 45.0 g of O₂. To determine the limiting reactant we have to convert those grams into moles of each compound. To do so we have to use their molar masses.

atomic mass of N = 14.01 amu

atomic mass of H = 1.01 amu

molar mass of N₂H₄ = 2 * 14.01 + 4 * 1.01

molar mass of N₂H₄ = 32.06 g/mol

moles of N₂H₄ = mass of N₂H₄ /(molar mass of N₂H₄)

moles of N₂H₄ = 30.0 g/(32.06 g/mol)

moles of N₂H₄ = 0.936 moles

atomic mass of O = 16.00 amu

molar mass of O₂ = 2 * 16.00

molar mass of O₂ = 32.00 g/mol

moles of O₂ = mass of O₂/(molar mass of O₂)

moles of O₂ = 45.0 g/(32.00 g/mol)

moles of O₂ = 1.41 moles

So we found that we mixed 0.936 moles of N₂H₄ with 1.41 moles of O₂. If we pay attention to the equation of the reaction:

N₂H₄ + O₂ ----> N₂ + 2 H₂O

We will see that the relationship or ratio between N₂H₄ and O₂ is 1 to 1 since their coefficients are 1. So 1 mol of N₂H₄ will react with 1 mol of O₂. We can use that ratio to find the limiting reactant and the one that is in excess. Let's take for example the 0.936 moles of N₂H₄.

theoretical moles of O₂ = 0.936 moles of N₂H₄ * 1 mol of O₂/(1 mol of N₂H₄)

theoretical moles of O₂ = 0.936 moles of O₂

As we said before, since both coefficients are 1, 0.936 moles of N₂H₄ will theoretically react with 0.936 moles of O₂. But we used 1.41 moles of O₂. So O₂ is in excess and N₂H₄ is the limiting reactant.

N₂H₄ ----> limiting reactant

O₂ --------> excess

Once we found the limiting reactant we can use it to find the moles of H₂O produced and finally the mass of it.

N₂H₄ + O₂ ----> N₂ + 2 H₂O

Looking again at the balanced equation we will see that 1 mol of N₂H₄ (when reacting with excess of O₂) will produce 2 moles of H₂O. We can use that relationship to find the number of moles of H₂O produced by 0.936 moles of N₂H₄.

moles of H₂O = 0.936 moles of N₂H₄ * 2 moles of H₂O/(1 mol of N₂H₄)

moles of H₂O = 1.872 moles

And finally we can use the molar mass of H₂O to find the answer to our problem.

atomic mass of H = 1.01 amu

atomic mass of O = 16.00 amu

molar mass of H₂O = 2 * 1.01 + 1 * 16.00

molar mass of H₂O = 18.02 g/mol

mass of H₂O = number of moles of H₂O * molar mass of H₂O

mass of H₂O = 1.872 moles * 18.02 g/mol

mass of H₂O = 33.7 g

Answers:

Equation: N₂H₄ + O₂ ----> N₂ + 2 H₂O

Limiting Reactant: N₂H₄

Excess: O₂

Grams of target product produced: 33.7 g of H₂O

User Eric Stein
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