For the reaction NH3+? O2 →? NO+? H2O ,
what is the maximum amount of NO
(30.0061 g/mol) which could be formed from
3.79 g of NH3 (17.0305 g/mol) and 18.63 g of O2 (31.9988 g/mol)?
Answer in units of g
-My first thought was to balance the equation, but I have no idea what to do after that.
This is a limiting reagent (LR) problem and the way you know that is that amounts are given for BOTH reactants.
Yes, balance the equation first. It would have been nice to do so and I could check the result.
1. Balance the equation.
2a. Convert 3/79 g N to mols. mols = grams/molar mass = ?
2b. Do the same and convert 18.63 g O2 to mols.
3a. Using the coefficients in the balanced equation, convert mols NH3 to mols of NO.
3b. Do the same and convert mols O2 to mols NO.
3c. It is likely that the two values for mols NO will not agree; the correct value to choose is ALWAYS the smaller value and the reagent responsible for that smaller value is the LR.
4. Convert the smaller value for mols NO to g NO. g NO = mols NO x molar mass NO = ?
Post your work if you get stuck.
To solve this problem, we need to follow a series of steps:
1. Write and balance the chemical equation:
NH3 + O2 → NO + H2O
2. Convert the given masses of NH3 and O2 to moles using the molar mass:
Molar mass of NH3 = 17.0305 g/mol
Molar mass of O2 = 31.9988 g/mol
moles of NH3 = 3.79 g / 17.0305 g/mol
moles of O2 = 18.63 g / 31.9988 g/mol
3. Determine the limiting reactant:
To find the limiting reactant, we compare the moles of the reactants to the stoichiometry of the reaction. The ratio of NH3 to NO is 1:1, and the ratio of O2 to NO is 5:4.
moles of NO from NH3 = moles of NH3
moles of NO from O2 = (moles of O2) * (4/5)
The reactant that produces less moles of NO is the limiting reactant.
4. Calculate the maximum moles of NO that can be formed:
We take the moles of the limiting reactant and multiply it by the stoichiometric ratio with NO.
moles of NO = minimum(moles of NO from NH3, moles of NO from O2)
5. Convert the moles of NO to grams using the molar mass:
mass of NO = moles of NO * molar mass of NO
Finally, substitute the values we found:
moles of NH3 = 3.79 g / 17.0305 g/mol = 0.2227 mol
moles of O2 = 18.63 g / 31.9988 g/mol = 0.5817 mol
moles of NO from NH3 = 0.2227 mol
moles of NO from O2 = 0.5817 mol * (4/5) = 0.4654 mol
moles of NO = minimum(0.2227 mol, 0.4654 mol) = 0.2227 mol
mass of NO = 0.2227 mol * 30.0061 g/mol = 6.6833 g
Therefore, the maximum amount of NO that can be formed is 6.6833 grams.