N2 + 3H2 + 2NH3 .

28.0 grams of N2 and 5.04 grams of H2 are reacted, producing 17.8 grams of NH3. What is the percent yield?

1. 62.9%
2. 84.0%
3. 52.4%
4. Not enough information is given.
5. 34.0%
6. 14.7%
7. 191%​

Answers

Answer 1

If 28.0 grams of N2 and 5.04 grams of H2 are reacted, producing 17.8 grams of NH3. the percent yield is 52.4%. Therefore, the answer is option 3,

To calculate the percent yield, we first need to calculate the theoretical yield of NH3 based on the amounts of N2 and H2 that were reacted.

The balanced chemical equation for the reaction is:

N2 + 3H2 → 2NH3

From the equation, we can see that 1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3. Therefore, the moles of N2 and H2 reacted can be calculated as:

moles of N2 = mass of N2 / molar mass of N2

= 28.0 g / 28.014 g/mol

= 0.999 mol

moles of H2 = mass of H2 / molar mass of H2

= 5.04 g / 2.016 g/mol

= 2.5 mol

Since N2 is the limiting reactant (its mole value is less than H2's mole value), we can calculate the amount of NH3 produced based on the mole ratio of N2 and NH3:

moles of NH3 = 2/1 x moles of N2

= 2/1 x 0.999 mol

= 1.998 mol

The theoretical yield of NH3 can be calculated from its mole value:

mass of NH3 = moles of NH3 x molar mass of NH3

= 1.998 mol x 17.031 g/mol

= 34.06 g

Now, we can calculate the percent yield by dividing the actual yield (17.8 g) by the theoretical yield (34.06 g) and multiplying by 100:

percent yield = actual yield / theoretical yield x 100%

= 17.8 g / 34.06 g x 100%

= 52.24%

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The molarity of a substance is simply defined as the mole of solute per unit litre of the solution. Mathematically, it is expressed as:

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