Which statement below is false when 10 g of nitrogen reacts with 5.0 g of hydrogen to produce ammonia?
N2 (g) + 3 H2 (g) → 2 NH3 (g)

Answers

Answer 1

Answer:

The theoretical yield of ammonia is 15 g.

Explanation:

The statements are:

2.8g of hydrogen are left over.

Hydrogen is the excess reactant.

Nitrogen is the limiting reactant.

The theoretical yield of ammonia is 15 g.

To solve this question we must find the moles of each reactant. Thus, we can find the limiting reactant. With limitinf reactant we can find the theoretical yield and the amount in excess of the reactant that is in excess:

Moles Nitrogen -28g/mol-

10.0g * (1mol / 28g) = 0.357 moles

Moles hydrogen -2g/mol-

5.0g * (1mol / 2g) = 2.5 moles

For a complete reaction of 0.357 moles of nitrogen are required:

0.357mol N2 * (3mol H2 / 1molN2) = 1.071 moles hydrogen.

As there are 2.5 moles of hydrogen, hydrogen is the excess reactant and nitrogen is the limiting reactant

The moles of hydrogen that are left over are:

2.5moles - 1.071moles = 1.429moles Hydrogen

The mass is:

1.429moles Hydrogen * (2g / mol) = 2.8g of hydrogen are left over

The moles of ammonia produced are:

0.357mol N2 * (2mol NH3 / 1molN2) = 0.714 moles of ammonia are produced (Theoretical moles)

The theoretical mass is (Molar mass NH3 = 17g/mol):

0.714 moles of ammonia * (17g / mol) =

12g of ammonia is the theoretical mass

The false statement is:

The theoretical yield of ammonia is 15 g.

Answer 2

The false statement, when 10 g of nitrogen reacts with 5.0 g of hydrogen to produce ammonia is that the theoretical yield of ammonia is 15g.

How moles is calculated?

Moles of any substance will be calculated as:

n = W/M, where

W = given mass

M = molar mass

Given chemical reaction is:
N₂(g) + 3 H₂(g) → 2 NH₃(g)

Moles of 10g of N₂ = 10/28 = 0.357 moles

Moles of 5g of H₂ = 5/2 = 2.5 moles

From the stoichiometry of the reaction:

1 mole of N₂ = react with 3 moles of H₂

0.357 moles of N₂ = react with 3×0.357=1.071 moles of H₂

Here nitrogen is the limiting reactant and hydrogen is excess reactant, and formation of ammonia depends on the limiting reactant.

1 mole of N₂ = react with 2 moles of NH₃

0.357 moles of N₂ = react with 2×0.357=0.714 moles of NH₃

Now we calculate the mass of ammonia from the given moles as:

M = (0.714) (17) = 12 g.

Hence, the false statement is that theoretical yield of ammonia is 15g.

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Answers

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Answers

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Answers

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Explanation:

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