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Mariana [72]
3 years ago
9

If 125.0g of nitrogen is reacted with 125.0g of hydrogen, what is the theoretical yield of the reaction? What is the excess reac

tant? What is the limiting reactant? Show your work.
Chemistry
1 answer:
MakcuM [25]3 years ago
5 0

Answer:

Hydrogen is the excess reactant

Nitrogen is the limiting reactant

151.6g is theoretical yield

Explanation:

The reaction of N₂ with H₂ to produce NH₃ is:

N₂ + 3H₂ → 2NH₃

To find theoretical yield we need to determine limiting reactant with the moles of each gas as follows:

Nitrogen -Molar mass: 28g/mol-

125.0g * (1mol / 28g) = 4.46 moles

Hydrogen -Molar mass: 2g/mol-

125.0g * (1mol / 2g) = 62.5 moles of hydrogen

For a complete reaction of 4.46 moles of N2 there are needed:

4.46 moles N2 * (3moles H2 / 1mol N2) = 13.38 moles of hydrogen

As there are 62.5 moles of hydrogen:

<h3>Hydrogen is the excess reactant</h3><h3>Nitrogen is the limiting reactant</h3><h3 />

With nitrogen, the limiting reactant, we determine theoretical moles (Assuming 100% of the reaction occurs) and theoretical yield (In mass):

4.46 moles N2 * (2moles NH3 / 1mol N2) = 8.92 moles of ammonia

As molar mass of ammonia is 17g/mol:

8.92 moles of ammonia * (17g/mol) =

<h3>151.6g is theoretical yield</h3>

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Consider the following reaction at constant P. Use the information here to determine the value of ΔSsurr at 355 K. Predict wheth
Elena-2011 [213]

Answer:

\Delta S_{surr} = + 0.32113\: kJ/K

Explanation:

Given: Entropy of surrounding: ΔSsurr = ?

Temperature: T= 355 K

The change in enthalpy of reaction: ΔH = -114 kJ

Pressure: P = constant

As we know, ΔH = -114 kJ ⇒ negative

Therefore, the given reaction is an exothermic reaction

Therefore, Entropy of surrounding at <em>constant pressure</em> is given by,

\Delta S_{surr} = \frac{-\Delta H}{T}

\therefore \Delta S_{surr} = -\left (\frac{-114 kJ}{355 K}  \right ) = + 0.32113\: kJ/K > 0

<u><em>In the given reaction:</em></u>

2NO(g) + O₂(g) → 2NO₂(g)

As, the number of moles of gaseous products is less than the number of moles of gaseous reactants.

\therefore \Delta S_{system} <  0

As we know, <em>for a spontaneous process, that the total entropy should be positive.</em>

\Delta S_{total} = \Delta S_{surr} + \Delta S_{system} > 0  

<u>Therefore, at the given temperature,</u>

  • if \Delta S_{surr} > \Delta S_{system} \Rightarrow \Delta S_{total} > 0 then the given reaction is spontaneous
  • if \Delta S_{surr} < \Delta S_{system} \Rightarrow \Delta S_{total} < 0 then the given reaction is non-spontaneous
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