Question

If 275.0g of nitrogen is reacted with 275.0g of hydrogen, what is the theoretical yield of the reaction? What is

the excess reactant? What is the limiting reactant? Show your work.

Answer 1

Theoretical yield of ammonia from 275.0g N₂ and 275.0g H₂ is 334.69g. Nitrogen is the limiting reactant, and hydrogen is the excess reactant.

Sure, let's go through the calculations step by step:

Step 1: Write the balanced chemical equation:

`\[ N_2 + 3H_2 \rightarrow 2NH_3 \]`

Step 2: Convert the masses of reactants to moles:

`\[ \text{Moles of } N_2 = \frac{275.0 \, \text{g}}{28.02 \, \text{g/mol}} \approx 9.821 \, \text{mol} \]`

`\[ \text{Moles of } H_2 = \frac{275.0 \, \text{g}}{2.02 \, \text{g/mol}} \approx 136.14 \, \text{mol} \]`

Step 3: Determine the mole ratio:

The balanced equation shows that 1 mole of
`\( N_2 \)` reacts with 3 moles of
`\( H_2 \).`

Step 4: Identify the limiting reactant:

`\[ \text{Moles of } NH_3 \text{ from } N_2 = 9.821 * 2 = 19.642 \, \text{mol} \]`

`\[ \text{Moles of } NH_3 \text{ from } H_2 = 136.14 * (2)/(3) \approx 90.76 \, \text{mol} \]`

The limiting reactant is
`\( N_2 \)`because it produces the smaller amount of
`\( NH_3 \) (19.642 mol).`

Step 5: Determine the theoretical yield:

`\[ \text{Theoretical yield} = 19.642 \, \text{mol} * \text{Molar mass of } NH_3 \]`

`The molar mass of \( NH_3 \) is approximately \( 17.03 \, \text{g/mol} \).\[ \text{Theoretical yield} = 19.642 \, \text{mol} * 17.03 \, \text{g/mol} \approx 334.69 \, \text{g} \]`

So, the theoretical yield is approximately
`\( 334.69 \, \text{g} \).`

Step 6: Identify the excess reactant:

The excess reactant is
`\( H_2 \)` because it is not completely consumed.

In summary:

- The theoretical yield of the reaction is approximately
`\( 334.69 \, \text{g} \).`

- The limiting reactant is
`\( N_2 \).`

- The excess reactant is
`\( H_2 \).`