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3 years ago

How many grams of nh3 can be produced from the reaction of 28 g of n2 and 25 g of h2

Chemistry

1 answer:

vfiekz [6]3 years ago

5 0

Answer:

34 g of NH₃ will be produced from the reaction of 28 g of N₂ with 25 g of H₂.

Explanation:

1) Balanced chemical equation

N₂ (g) + 3H₂ (g) → 2NH₃(g)

2) Stoichiometric (theoretical ) mole ratios

1 mol N₂ (g) : 3mol H₂ (g) : 2 mol NH₃(g)

3) Number of moles of each reactant

number of moles = mass in grams / atomic mass

number of moles of N₂ = 28 g / 28 g/mol = 1 mol

number of moles of H₂: 25 g / 2 g/mol = 12.5 mol

4) Limiting reactant

Since the stoichiometry states that 1 mol of N₂ reacts with 3 moles of H₂, the given mass of N₂ will react completely with the given amount of H₂, and the calculations must be done with the 28 g (1 mol) of N₂ as the limiting reactant.

5) Yield

Set the proportion with the mole ratios:

1 mol H₂ / 2 mol NH₃ = 1 mol H₂ / x ⇒ x = 2 mol NH₃

6) Convert to grams

mass in grams = number of moles × molar mass = 2 mol × 17 g/mol = 34 g.

Answer: the reaction of 28 g of N₂ with 25 g of H₂ will produce 34 g of NH₃

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3 years ago

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2 years ago

What factors affect the dynamic state of equilibrium in a chemical reaction and how?

yanalaym [24]

Answer:

Only changes in temperature will influence the equilibrium constant $K_c$ . The system will shift in response to certain external shocks. At the new equilibrium $Q$ will still be equal to $K_c$ , but the final concentrations will be different.

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External changes in the relative concentration of the products and reactants.

For some reversible reactions that involve gases:

Changes in pressure due to volume changes.

Catalysts do not influence the value of $Q$ . See explanation.

Explanation:

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Similar to the rate constant, the equilibrium constant $K_c$ depends only on:

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The reversible reaction is in a dynamic equilibrium when the rate of the forward reaction is equal to the rate of the backward reaction. Reactants are constantly converted to products; products are constantly converted back to reactants. However, at equilibrium $Q = K_c$ the two processes balance each other. The concentration of each species will stay the same.

Factors that alter the rate of one reaction more than the other will disrupt the equilibrium. These factors shall change the rate of successful collisions and hence the reaction rate.

Changes in concentration influence the number of particles per unit space.
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For reactions that involve gases,

Changing the volume of the container will change the concentration of gases and change the reaction rate.

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3 years ago

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Answer:

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