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ELEN [110]
2 years ago
12

1N2 + 3H2 -->

Chemistry
1 answer:
Hunter-Best [27]2 years ago
5 0

Answer:

28.23 g NH₃

Explanation:

The balanced chemical equation is:

N₂(g) + 3 H₂(g) → 2 NH₃(g)

Thus, 1 mol of N₂ reacts with 2 moles of H₂ to produce 2 moles of NH₃. We convert the moles to mass (in grams) by using the molecular weight (MW) of each compound:

MW(N₂) = 2 x 14 g/mol = 28 g/mol

mass N₂= 1 mol x 28 g/mol = 28 g

MW(H₂) = 2 x 1 g/mol = 2 g/mol

mass H₂ = 3 mol x 2 g/mol = 6 g

MW(NH₃) = 14 g/mol + (3 x 1 g/mol) = 17 g/mol

mass NH₃= 2 moles x 17 g/mol = 34 g

Now, we have to figure out which is the limiting reactant. For this, we know that the stoichiometric ratio is 28 g N₂/6 g H₂. If we have 36.85 g of H₂, we need the following mass of N₂:

36.85 g H₂ x 28 g N₂/6 g H₂ = 171.97 g N₂

We have 23.15 g N₂ and we need 171.97 g. So, we have lesser N₂ than we need. Thus, the limiting reactant is N₂.

Now, we calculate the product (NH₃) by using the stoichiometric ratio 34 g NH₃/28 g N₂, with the mass of N₂ we have:

23.25 g N₂ x 34 g NH₃/28 g N₂ = 28.23 g NH₃

Therefore, the maximum amount of NH₃ that can be produced is 28.23 grams.

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Read 2 more answers
A model used for the yield Y of an agricultural crop as a function of the nitrogen level N in the soil (measured in appropriate
lisov135 [29]

Answer:

  • It is 3.

Explanation:

The model written correctly is:

  • Y=\frac{kN}{9+N^2}

This is a mathematical question, instead of a chemistry question, and you should use calculus to find the nitrogen level that gives the best yield, since this is an optimization problem.

The best yield is the maximum yield, and the maximum, provided that it exists, is found using the first derivative and making it equal to zero: Y' = 0

To find Y' you must use the quotient rule.

Y'=\frac{(kN)'(9+N^2)- (kN)(9+N^2)'}{(9+N^2)^2}\\ \\Y'=\frac{k(9+N^2)-kN(2N)}{(9+N^2)^2}\\ \\Y'=\frac{9k + kN^2 - 2kN^2}{(9 + N^2)^2}=\frac{9k-kN^2}{(9+N^2)^2}

Now make Y' = 0

  • The denominator is never equal to zero, because it is always positive and greater than 9.

  • Make the numerator equal to zero:

         9k - kN² = 0

  • Factor: k (9 - N²) = 0

  • Since k is a positve constant, it is not equal to zero, and the other factor, 9 - N², must be equal to zero:

         9 - N² = 0 ⇒ (3 - N) (3 + N) = 0

         ⇒ 3 - N = 0 or 3 + N = 0 ⇒ N = 3 or N = -3.

Since N is nitrogen level, it cannot be negative and the only valid answer is N = 3.

You can prove that it is a maximum (instead of a minimum) finding the second derivative or testing some points around 3 (e.g. 2.5 and 3.5).

8 0
3 years ago
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