A sample of 8.5 g nh3 on oxidation produces 4.5 g of no. calculate the percent yield. reaction: 4 nh3 5 o2 → 4 no 6 h2o

1 answer:

5 0

<span>30.% Determine the molar mass of NH3 and NO Atomic weight nitrogen = 14.0067 Atomic weight hydrogen = 1.00794 Atomic weight oxygen = 15.999 Molar mass NH3 = 14.0067 + 3 * 1.00794 = 17.03052 g/mol Molar mass NO = 14.0067 + 15.999 = 30.0057 g/mol Moles NH3 = 8.5 / 17.03052 = 0.499103962 mol Moles NO = 4.5 / 30.0057 = 0.149971505 mol Looking at the balanced equation, for every mole of NH3 consumed, you should get one mole of NO. So if we had 100% yield, we should have 0.499103962 moles of NO. But we don't. The percent yield is a simple matter of division by what we did get by what we should get. So 0.149971505 / 0.499103962 = 0.300481497 = 30.0481497% Rounding to 2 significant figures gives 30.% yield.</span>

You might be interested in

jin has four colorful glass bottles on his windowsill. one is red, one is violet, one is green, and one is yellow. the light tha

Over [174]

The red bottle would have the lowest frequency because red light has the longest wavelengths. The light passing through the violet would have the highest frequency because its wavelengths are the shortest.

4 0

2 years ago

Read 2 more answers

A solution is made by dissolving 58.125 g of sample of an unknown, nonelectrolyte compound in water. The mass of the solution is

e-lub [12.9K]

Answer:

molecular weight (Mb) = 0.42 g/mol

Explanation:

mass sample (solute) (wb) = 58.125 g

mass sln = 750.0 g = mass solute + mass solvent

∴ solute (b) unknown nonelectrolyte compound

∴ solvent (a): water

⇒ mb = mol solute/Kg solvent (nb/wa)

boiling point:

ΔT = K*mb = 100.220°C ≅ 373.22 K

∴ K water = 1.86 K.Kg/mol

⇒ Mb = ? (molecular weight) (wb/nb)

⇒ mb = ΔT / K

⇒ mb = (373.22 K) / (1.86 K.Kg/mol)

⇒ mb = 200.656 mol/Kg

∴ mass solvent = 750.0 g - 58.125 g = 691.875 g = 0.692 Kg

moles solute:

⇒ nb = (200.656 mol/Kg)*(0.692 Kg) = 138.83 mol solute

molecular weight:

⇒ Mb = (58.125 g)/(138.83 mol) = 0.42 g/mol

8 0

1 year ago

Nitrogen gas can be prepared by passing gaseous ammonia over solid copper (II) oxide at high temperatures. If 18.1 g of Nh3 is r

Scilla [17]

I first converted the given grams of the reactants into moles, and then divided the moles by the coefficients in front of each of the reactant. The result with the smallest value will be the limiting reactant, and the value of CuO was the smallest, so it's the limiting reactant.

After figuring out which reactant is the limiting one, I took their given grams and converted it into moles, the divided it by the ratio of N2 to CuO (it's in the equation) to obtain the moles of N2, and then multiply it with the molar mass of N2 to get its mass in grams.