Question

I need help with this, please :00000

Answer 1

theoretical yield of the reaction is 121.38 g of NH₃ (ammonia)

limiting reactant is N₂ (nitrogen)

excess reactant is H₂ (hydrogen)

Explanation:

We have the following chemical reaction:

N₂ + 3 H₂ → 2 NH₃

Now we calculate the number of moles of each reactant:

number of moles = mass / molar weight

number of moles of N₂ = 100 / 28 = 3.57 moles

number of moles of H₂ = 100 / 2 = 50 moles

From the chemical reaction we see that 3 moles of H₂ are reacting with 1 moles of N₂, so 50 moles of H₂ are reacting with 16.66 moles of N₂ but we only have 3.57 moles of  N₂ available, so the limiting reactant will be N₂ and the excess reactant will be H₂.

Knowing the chemical reaction and the limiting reactant we devise the following reasoning:

if          1 mole of N₂ produce 2 moles of NH₃

then    3.57 moles of N₂ produce X moles of NH₃

X = (3.57 × 2) / 1 = 7.14 moles of NH₃

mass = number of moles × molar weight

mass of NH₃ = 7.14 × 17 = 121.38 g

theoretical yield of the reaction is 121.38 g of NH₃

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Answer 2

Answer:

Use stoichiometry to convert 100.0 g of N2 to g of NH3.  This calculated mass will be the theoretical yield.  34.0 g of NH3 obtained is called the actual yield.  %-Yield = (actual yield / theoretical yield) x 100%.

For the calculation, molar mass of N2 is 28 g/mol, molar mass of NH3 is 17 g/mol, and mole ratio of N2 to NH3 from the balanced chemical reaction is 1:2.

100.0 g N2 x (1 mol N2 / 28 g N2) x (2 mol NH3 / 1 mol N2) x (17 g NH3 / 1 mol NH3) = 121.4 g NH3

%-Yield = (Actual yield / Theoretical yield) x 100%

           = (34.0 g / 121.4 g) x 100%

           = 28.0%

Explanation: