for the reaction: 4NH3+5O2=4NO+6H2O; 15.0g of NH3 and 27.5g of O2 were reacted.
a). which is the limiting reactant?
b).how many moles of NO are formed from the limiting reactant in the above reaction.
i know how to do the moles but i couldn't figure out which one is the limiting reactant. to do the moles first i figured out the molecular weight for NO(30.01). but i don't understand what the mass in grams will be.
Just think of limiting reagent problems as being two problems. Work them separately. First convert g NH3 and g O2 to mols. Then make the two problems out of it.
How many mols NO will be formed with ?? mols NH3 AND ALL THE O2 NEEDED. Get a mols NO for that.
Second problem. How many mols NO will be formed with xx mols O2 AND ALL THE NH3 NEEDED?
The SMALLER number of mols product formed will ALWAYS be the correct answer and the reactant producing that number will be the limiting reagent. There are quicker ways of doing this but this works well and it's easy to explain. You can work on quicker ways of doing it after you get the hang of it.
oh, i think i get it. thank you.
What is a balance chemical equation for nitric acid is a compound of acid rain that forms when gaseous nitrogen dioxide pollutant reacts with gaseous oxygen and liquid water to format aqueous nitric acid
Given the combustion reaction of methane below, determine the (a) limiting reactant and (b) mass
of CO2 formed if 17.0 g of CH4 reacts with 48.0 g of O2
To determine the limiting reactant in a chemical reaction, you need to compare the moles of each reactant to their stoichiometric coefficients (the numbers in front of each reactant in the balanced equation).
a) Let's calculate the moles of NH3 and O2 based on their given masses.
Molar mass of NH3 (ammonia) = 14.01 g/mol + 3(1.01 g/mol) = 17.03 g/mol
Moles of NH3 = mass of NH3 / molar mass of NH3
= 15.0 g / 17.03 g/mol
≈ 0.881 mol
Molar mass of O2 (oxygen gas) = 2(16.00 g/mol) = 32.00 g/mol
Moles of O2 = mass of O2 / molar mass of O2
= 27.5 g / 32.00 g/mol
≈ 0.859 mol
Now, compare the ratio of moles of NH3 to O2 based on their stoichiometric coefficients:
4 NH3 : 5 O2
From the moles calculated above, we can see that there are fewer moles of O2 than NH3. Therefore, O2 is the limiting reactant in this case.
b) To calculate the moles of NO formed from the limiting reactant, we need to use the stoichiometry of the balanced equation.
From the balanced equation: 4 NH3 + 5 O2 → 4 NO + 6 H2O
The stoichiometric coefficient of NO is 4, which means for every 4 moles of NH3 reacted, 4 moles of NO are formed.
Since O2 is the limiting reactant, we can use the ratio of O2 to NO to calculate the moles of NO:
5 O2 : 4 NO
From the moles calculated for O2, we can see that there are 0.859 moles.
Moles of NO = moles of O2 × (4 moles of NO / 5 moles of O2)
= 0.859 mol × (4/5)
≈ 0.687 mol
Therefore, approximately 0.687 moles of NO are formed from the limiting reactant, which is O2, in the given reaction.