I need the answer for each empty box

Ammonia is produced by the following reaction. 3H2(g) N2(g) Right arrow. 2NH3(g) When 7. 00 g of hydrogen react with 70. 0 g of

harkovskaia [24]

In the ammonia production process given by the reaction 3H₂(g) + N₂(g) → 2NH₃(g), when 7.00 g of hydrogen react with 70.0 g of nitrogen, hydrogen is considered the limiting reactant because 7.5 moles of hydrogen would be needed to consume the available nitrogen (option 1).

The reaction is the following:

3H₂(g) + N₂(g) → 2NH₃(g) (1)

To know why hydrogen is considered the limiting reactant, we need to calculate the number of moles of nitrogen and hydrogen with the following equation:

$n = \frac{m}{M}$

Where:

m: is the mass

M: is the molar mass

For hydrogen we have:

$n_{H_{2}} = \frac{m}{M} = \frac{7.00 g}{2.016 g/mol} = 3.47 \:moles$

And for nitrogen:

$n_{N_{2}} = \frac{m}{M} = \frac{70.0 g}{28.013 g/mol} = 2.50 \:moles$

We can see in reaction (1) that 3 moles of hydrogen react with 1 mol of nitrogen, so the number of hydrogen moles needed to react nitrogen is:

$n_{H_{2}} = \frac{3\:moles\:H_{2}}{1\:moles\:N_{2}}*n_{N_{2}} = \frac{3\:moles\:H_{2}}{1\:moles\:N_{2}}*2.50 \:moles = 7.50 \:moles$

Since we have 3.47 moles of hydrogen and we need 7.50 moles to react with all the mass of nitrogen, the limiting reactant is hydrogen.

We can find the number of ammonia moles produced with the limiting reactant (hydrogen) konwing that 3 moles of hydrogen produces 2 moles of ammonia, so:

$n_{NH_{3}} = \frac{2\:moles\:NH_{3}}{3\:moles\:H_{2}}*n_{H_{2}} = \frac{2\:moles\:NH_{3}}{3\:moles\:H_{2}}*3.47 \:moles = 2.31 \:moles$

Hence, hydrogen would produce 2.31 moles of ammonia.

Therefore, hydrogen is the limiting reactant because 7.5 moles of hydrogen would be needed to consume the available nitrogen (option 1).

Find more about limiting reactants here:

brainly.com/question/2948214?referrer=searchResults

I hope it helps you!

6 0

3 years ago

A type of emergency apparatus that can be used where oxygen may be limited or where the air might be poisoned is based on the fo

Bess [88]

Answer:

so I don't know how to answer that

8 0

4 years ago

Select the correct answer.

vitfil [10]

Answer:

The answer is B

Explanation:

6 0

4 years ago

Read 2 more answers

How did uranium become present in Earth's upper crust? Choose one: A. It was delivered by comets that crashed into Earth's surfa

Luda [366]

Answer:

Option A. It was delivered by comets that crashed into Earth's surface.

Explanation:

Uranium (U) is a chemical element with atomic number 92.

For many years, a large number of scientists have been studying the abundance and origin of the isotopes of uranium in Earth. According to some theories, the Earth's uranium was produced in one or more supernovae (an explosive brightening of a star), in wich, the main process consists in the rapid capture of neutrons by seed nuclei at great rates. Another theory proposes that uranium is created during the merger of two neutron stars (neutron stars are very dense), because, when such dense bodies come closer together the gravitational force cause them to merge, producing huge amounts of hevy metals like uranium.

Many analyses have been made of the uranium in rocks of the Earth. These measurements shows that the abundance of uranium is bigger in the crust and upper mantle of the Earth.

So, knowing that Earth's uranium was produced through one of these processes, the best answer is option A, the uranium was delivered by comets that crashed into Earth's surface.

Have a nice day!

5 0

4 years ago

A container initially holds 5.67 x 10^-2 mol of propane and has a volume of V1. The volume of the container was increased by add

Brut [27]

Answer:

Initial volume of the container (V1) = 1.27 L (Approx)

Explanation:

Given:

Number of mol (n1) = 5.67 x 10⁻²

Number of mol (n2) = (5.67 +2.95) x 10⁻² = 8.62 x 10⁻²

New volume (V2) = 1.93 L

Find:

Initial volume of the container (V1)

Computation:

Using Avogadro's law

V1 / n1 = V2 / n2

V1 / 5.67 x 10⁻² = 1.93 / 8.62 x 10⁻²

V1 = 10.9431 / 8.62

Initial volume of the container (V1) = 1.2695

Initial volume of the container (V1) = 1.27 L (Approx)

5 0

3 years ago