In an experiment, 6.25 g of NH3 are allowed to react with 8.50 g of O2. The chemical equation for the reaction is 4NH3 + 5O2 → 4NO + 6 H2O. How many grams of water are produced? Which reactant limits the production of water? Which reactant is in excess?
To determine the number of grams of water produced in the reaction, we first need to figure out the limiting reactant. The limiting reactant is the reactant that is entirely consumed, and it determines the maximum amount of product that can be formed.
To find the limiting reactant, we compare the amount of each reactant to its stoichiometric coefficients in the balanced equation.
Let's start by converting the given masses of NH3 and O2 to moles:
Molar mass of NH3 (ammonia) = 14 g/mol + 3(1 g/mol) = 17 g/mol
Molar mass of O2 (oxygen) = 2(16 g/mol) = 32 g/mol
Number of moles of NH3 = 6.25 g / 17 g/mol = 0.3676 mol (approximately)
Number of moles of O2 = 8.50 g / 32 g/mol = 0.2656 mol (approximately)
Now, let's determine the limiting reactant by comparing the moles of NH3 and O2 to their stoichiometric coefficients in the balanced equation.
From the balanced equation, we see that the ratio of NH3 to H2O is 4:6, which simplifies to 2:3. Therefore, the number of moles of H2O produced is:
0.3676 mol (NH3) x (3 mol H2O / 2 mol NH3) = 0.5514 mol of H2O from NH3
From the balanced equation, we see that the ratio of O2 to H2O is 5:6. Therefore, the number of moles of H2O produced is:
0.2656 mol (O2) x (6 mol H2O / 5 mol O2) = 0.3187 mol of H2O from O2
We can see that the moles of H2O produced from NH3 (0.5514 mol) are greater than the moles produced from O2 (0.3187 mol).
Therefore, NH3 is the limiting reactant in this reaction.
To determine the grams of water produced, we need to know the molar mass of water:
Molar mass of H2O = 2(1 g/mol) + 16 g/mol = 18 g/mol
Now, we can calculate the mass of water produced using the moles of water and the molar mass of water:
Mass of H2O produced = 0.5514 mol x 18 g/mol = 9.9272 g (approximately)
Therefore, approximately 9.93 grams of water are produced in this reaction.
In this reaction, NH3 is the limiting reactant, which means it is entirely consumed before the reaction can proceed further. O2 is the excess reactant, meaning that there is some amount left over after the reaction is complete.
A limiting reagent (LR) and excess reagent (ER) problem. You know that because amounts are given for BOTH reactants; besides, the problem tell you.
Convert 6.25 g NH3 to mols. mols = grams/molar mass
Do the same for O2.
Using the coefficients in the balanced equation, convert mols NH3 to mols H2O.
Do the same and convert mols O2 to mols H2O.
It is likely that thae two values for mols H2O will not agree; the correct value in LR problems is ALWAYS the smaller number AND the reagent responsible for that lower number is the LR. The other reagent is the ER.
Now convert mols H2O to grams. g = mols x molar mass.