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