Amanda L.

asked • 05/26/21

Chemistry question please if anyone can help I really need it

___NH3+___O2----> ____ NO +____H2O

In an experiment 5.03g of NH3 are allowed to react with 8.78g of O2 .

  1. Write the formula of the limiting reagent

ANSWER ____

  1. 2 How many grams of NO are formed ? Answer___ grams
  2. 3. How much of the excess reactant in grams remains after the reaction? Answer ___grams



Geoffrey C.

Hi Amanda, I think I can help you with this one. So, first you have a chemical equation that needs balancing. start by counting the number of each atom on the left and right side of the equations and place coefficients in front of each chemical so the the starting material (SM) is balanced by the products . make a table if it helps......Close but not balanced as its given. so clearly we see that the hydrogens are causing the trouble, so how can we fix it? N H O N H O 1 3 2 1 2 2
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05/26/21

Geoffrey C.

if we put a 2 in front of the NH# and put a 3 in front of H20 then both will have 6 hydrogens; like this 2NH3 +O2 = NO + 3H20, perfect, right? 6 hydrogens on each side!! but we must check all of the atoms and recount them all. so 2N, 6H, and 2O on left; 1N, 6H, and 4O....hmmmmm looks like we need to double the oxygen amount of the left side....so double it too. 2NH3 +2O2 = NO + 3H20 and we must double the nitrogen on the right....double it!!!! 2NH3 +2O2 = 2NO + 3H20 and now we have 2 nitrogen on each side! BUT we messed with oxygen equilibrium. So now what? we fix it.......4 on the left and 5 on the right.....welp, on the left we could have 4O2 for a total of 8 but if we just double the numbers (coefficients) on the right we get .........= 4NO +6H2O......10 Oxygen!!!! Hooray I know how to make the SM-side (left) have 10 oxygens too; its easy; 2NH3 +5O2 = 4NO + 6H20 we are so close 10 oxygen each side, 2 N left, 4 N right......ok lets double the 2NH3 to get 4NH3 +5O2 = 4NO + 6H20. NOW count the atoms on each side. 4N, (4x3H and 6x2H) = 12H and 10 O on each side.....now its balanced.
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05/26/21

Geoffrey C.

for the next part we need the molecular weights (MW) for each compound and then we need to convert the amount of SM (5.03 g NH3 and 8.78 g O2) into to mols. this part isn't so hard once you look up the MW. So 17.03 g NH3/mol, 32.00 g O2/mol, 30.01 g NO/mol and 18.02 g H2O/mol. all units of MW are g/mol here. NOW....the chemical equation tells us that we could take 4 mol NH3 and 5 mol of O2 and get 4 mol NO and 6 mol H2O but we weren't given reagents in mols, they gave us grams!!!!!! so convert, and I tell students to follow the units. here we go; 5.03 g NH3 x 1 mol/17.03 g NH3 (the MW is upside-down) = 5.03 g/17.03 mol = 0.295 mols NH3, next up 8.78 g O2 x 1 mol/32.00 = 0.274 mol O2. Ok now we ask our selves if ammonia and oxygen react in a 4 to 5 ratio (from the equation) do we have enough O2 to react all the ammonia? well it seem as thought we have less oxygen (in mols) than ammonia. the equation says we need MORE O2 (in mols) than NH3 to fully react everything, clearly we dont have enough oxygen and it will serve as the limiting reagent. Here's the math to prove it. start with the mols of limiting reagent and divide by its coefficient ort 0.274/5 = 0.0548 which is what we call 1 equivalent or 1 eq. so now we need 4 eq NH3 and we need 5 eq O2 so 5x 0.0548 = 0.274 mol O2 (which we have but just barely) and 0.0548 x 4 = 0.2192 mols NH3 (we know that we actually have 0.295 mol NH2 from above) so it looks like we have way too much NH3 and not enough O2 (0.295 mol NH3 - 0.2192 mol NH3 = 0.076 mols excess NH3) now they want to know how much NO is formed. its simple, we know from the equation that we need 4 eq. or 4x 0.0548 which comes out to 0.2192 (the same amount of mols of NH3 needed to react the 5O2). They ask for grams though so......0.2192 mols x 30.01 g NO/mol = 6.58 g NO are formed. note that the MW of NO is higher than that on NH3. if we have the same number of mols of each, as we do here, we should end up with more GRAMS of NO than of NH3.....check! and finally how much excess reagent were we given. well we just did that above, ill copy and paste it: (0.295 mol NH3 given - 0.2192 mol NH3 needed = 0.076 mols excess NH3) and we can easily convert 0.076 mols NH3 just multiply by ammonia's MW or 0.076 mols x 17.03 g NH3/mol = 1.29 grams of excess ammonia given to us......so wasteful, they should hire a real chemist, like perhaps You!, to get the job done without destroying the planet. Anyhow contact me here if you need more help. my tutoring sessions are NOT so wordy, and with a white board I can spend time actually teaching you this stuff. I know its a lot but if your still confused ------> [email protected] or message me here. GL :)
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05/26/21

J.R. S.

tutor
@Geoffrey C. It would be better if you posted your answer as an "answer" instead of in the "comment" section. Just a suggestion.
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05/26/21

1 Expert Answer

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J.R. S. answered • 05/26/21

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Ph.D. in Biochemistry--University Professor--Chemistry Tutor
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4NH3 + 5O2 ==> 4NO + 6H2O ... balanced equation


1). To find limiting reactant, simply divide moles of each reactant by its coefficient in the balanced equation.

For NH3 we have 5.03 g x 1 mol NH3 / 17 g = 0.296 moles (÷ 4 -> ~0.074)

For O2 we have 8.78 g x 1 mol O2 / 32 g = 0.274 moles (÷ 5 -> ~0.055)

In this case...O2 is limiting


2). 0.274 mols O2 x 4 mols NO / 5 mols O2 x 30 g NO / mol = 6.58 g NO formed


3). To find amount of NH3 (the excess reactant), we will first find amount NH3 used up and subtract it from amount originally present.

NH3 used = 0.274 mols O2 x 4 mols NH3 / 5 mols O2 = 0.219 mols NH3 used

moles originally present = 0.296 (see calculation above in #1)

NH3 left after the reaction = 0.296 mols - 0.219 mols = 0.077 mols x 17 g /mol = 1.31 g left over

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Amanda L.

Thank you so much
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05/26/21

J.R. S.

tutor
You're very welcome.
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05/26/21

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