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Tresset [83]
3 years ago
15

Under appropriate conditions, nitrogen and hydrogen undergo a combination reaction to yield ammonia: N2(g)+3H2(g)--->2NH3(g).

Chemistry
1 answer:
lora16 [44]3 years ago
8 0

Answer:

The answer to your question is:        14 g of N₂

Explanation:

Data

MW N2 = 14

MW H2 = 1

MW NH3 = 17

                                  N₂(g)  +   3H₂(g)    --->   2NH₃(g).

                                 28g           6g                  34g

               

                             28 g of N₂ --------------------  6 g of H₂

                               x              --------------------  3 g of H₂

                               x = (3 x 28) / 6

                               x = 14 g of N₂

                               

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Given the following data:
bagirrra123 [75]

176.0 \; \text{kJ} \cdot \text{mol}^{-1}

As long as the equation in question can be expressed as the sum of the three equations with known enthalpy change, its \Delta H can be determined with the Hess's Law. The key is to find the appropriate coefficient for each of the given equations.

Let the three equations with \Delta H given be denoted as (1), (2), (3), and the last equation (4). Let a, b, and c be letters such that a \times (1) + b \times (2) + c \times (3) = (4). This relationship shall hold for all chemicals involved.

There are three unknowns; it would thus take at least three equations to find their values. Species present on both sides of the equation would cancel out. Thus, let coefficients on the reactant side be positive and those on the product side be negative, such that duplicates would cancel out arithmetically. For instance, 3 + (-1) = 2 shall resemble the number of \text{H}_2 left on the product side when the second equation is directly added to the third. Similarly

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a = -1/2\\b = 1/2\\c = -1/2 and

-\frac{1}{2} \times (1) + \frac{1}{2} \times (2) - \frac{1}{2} \times (3)= (4)

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When FeC13 is ignited in an atmosphere of pure oxygen, this reaction takes place. 4FeCl3(sJ 30lgJ ~ 2F~0 (sJ 6Cl2(gJ If 3.00 mol
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Solution : Given,

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Limiting reagent : It is defined as the reactants completely used up in the reaction.

Now we have to calculate the limiting and excess reagent.

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