For the reaction
? N2H4+? O2 →? N2+? H2O ,
what is the maximum amount of N2
(28.0134 g/mol) which could be formed from
14.75 g of N2H4 (32.0452 g/mol) and 4.31 g of O2 (31.9988 g/mol)?
Answer in units of g
another limiting reagent problem.
To determine the maximum amount of N2 that can be formed from the given amounts of N2H4 and O2, we need to follow these steps:
1. Write and balance the chemical equation:
N2H4 + O2 → N2 + H2O
2. Calculate the moles of N2H4 and O2:
Moles of N2H4 = mass of N2H4 / molar mass of N2H4
Moles of O2 = mass of O2 / molar mass of O2
Moles of N2H4 = 14.75 g / 32.0452 g/mol
Moles of O2 = 4.31 g / 31.9988 g/mol
3. Determine the limiting reactant:
The limiting reactant is the one that is completely consumed in the reaction and determines the maximum amount of product that can be formed. To find the limiting reactant, compare the mole ratios of N2H4 and O2 to N2 in the balanced equation. The reactant with the smaller mole ratio is the limiting reactant.
From the balanced equation, the mole ratios are:
N2H4 : N2 = 1 : 1
O2 : N2 = 1 : 1
Since both mole ratios are the same, the limiting reactant will be the reactant that has the smaller number of moles.
4. Calculate the moles of N2 that can be formed:
The limiting reactant will react with another reactant according to the balanced equation. The coefficient of the limiting reactant represents the mole ratio between that reactant and the product (N2).
Since both N2H4 and O2 have a 1:1 mole ratio with N2, the limiting reactant will produce the same number of moles of N2.
Moles of N2 = Moles of limiting reactant
5. Convert moles of N2 to grams:
Grams of N2 = Moles of N2 × molar mass of N2
Moles of N2 = smaller number of moles from step 3
Molar mass of N2 = 28.0134 g/mol
Now, you can follow these steps to calculate the maximum amount of N2 that could be formed from the given amounts of N2H4 and O2.