Question

In the synthesis of ammonia, 42 g of nitrogen gas is reacted with 8 g of hydrogen gas. Mass of ammonia produced in the process is​

Answer 1

Final Answer:

The mass of ammonia produced in the process is 51 grams.

Detailed Explanation:

To determine the mass of ammonia produced in the synthesis of ammonia, we need to calculate the limiting reactant and then use stoichiometry to find the corresponding mass of ammonia.

First, let's determine the limiting reactant by comparing the amounts of nitrogen gas (N₂) and hydrogen gas (H₂) used. We will calculate the number of moles for each reactant using their molar masses:

Molar mass of nitrogen gas (N₂) = 28 g/mol

Molar mass of hydrogen gas (H₂) = 2 g/mol

Number of moles of nitrogen gas (N₂):

n(N₂) = mass(N₂) / molar mass(N₂) = 42 g / 28 g/mol = 1.5 mol

Number of moles of hydrogen gas (H₂):

n(H₂) = mass(H₂) / molar mass(H₂) = 8 g / 2 g/mol = 4 mol

The balanced equation for the synthesis of ammonia (NH₃) is:

N₂ + 3H₂ → 2NH₃

From the balanced equation, we can see that the stoichiometric ratio between nitrogen gas and ammonia is 1:2. Therefore, for every mole of nitrogen gas, we expect to produce 2 moles of ammonia.

Since we have 1.5 moles of nitrogen gas, the maximum number of moles of ammonia that can be produced is 2 times that:

n(NH₃) = 2 × n(N₂) = 2 × 1.5 mol = 3 mol

Now, let's calculate the mass of ammonia produced using its molar mass:

Molar mass of ammonia (NH₃) = 17 g/mol

Mass of ammonia produced:

mass(NH₃) = n(NH₃) × molar mass(NH₃) = 3 mol × 17 g/mol = 51 g

Therefore, the mass of ammonia produced in the process is 51 grams.

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Answer 2

Above given question is incomplete and complete question can be read as:

In the synthesis of ammonia, 42 g of nitrogen gas is reacted with 8 g of hydrogen gas. Mass of ammonia produced in the process is

1) 50 g

2) 36.5 g

3) 45.3 g

4) 48 g

Answer:

The option 1) 50 g is the final answer.

Explanation:

To determine the mass of ammonia produced in the synthesis of ammonia from nitrogen gas and hydrogen gas, we need to consider the balanced chemical equation and the stoichiometry of the reaction.

The balanced chemical equation for the synthesis of ammonia is:

N₂ + 3H₂ → 2NH₃

According to the given information, we have 42 g of nitrogen gas and 8 g of hydrogen gas.

To determine the limiting reactant, we need to calculate the moles of each reactant.

The molar mass of nitrogen gas is 28 g/mol, and the molar mass of hydrogen gas is 2 g/mol.

Moles of nitrogen gas = 42 g / 28 g/mol = 1.5 mol

Moles of hydrogen gas = 8 g / 2 g/mol = 4 mol

From the balanced equation, we can see that 1 mole of nitrogen reacts with 3 moles of hydrogen to produce 2 moles of ammonia.

Since the molar ratio of nitrogen to ammonia is 1:2, the maximum amount of ammonia that can be produced is determined by the limiting reactant, which is nitrogen in this case.

Moles of ammonia produced  = $\frac{(1.5 mol (moles of nitrogen) . 2 mol (moles of ammonia) }{1 mol (moles of nitrogen)} = 3$

To find the mass of ammonia produced, we multiply the moles of ammonia by its molar mass.

Mass of ammonia produced = 3 mol × 17 g/mol (molar mass of ammonia) = 51 g

51 ≈ 50

Therefore, the option 1) 50 g is the final answer which is close to 51 g.

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