Question

Ammonia and oxygen react to form nitrogen and water.

4NH3(g) + 302(g) → 2N2+ 6H2O(g)
a. How many grams of O2 are needed to react with 13.6 g
of NH3?
b. How many grams of N2 can be produced when 6.50 g of O2
reacts?
c. How many grams of H2O are formed from the reaction of
of NH3?
34.0 g

Answer 1

The questions involve using stoichiometry to convert grams of reactants to grams of products for the chemical reaction of ammonia and oxygen, which form nitrogen and water. It requires an understanding of molar masses, mole-to-gram conversions, and the use of molar ratios from the balanced equation. a. 32 g/mol. b. 28 g/mol. c. 18 g/mol.

Step-by-step explanation:

To solve the questions related to the reaction of ammonia (NH₃) and oxygen (O₂), we first need to interpret the balanced chemical equation:

4NH₃(g) + 3O₂(g) → 2N₂(g) + 6H₂O(g)

• a. To find how many grams of O₂ is needed to react with 13.6 g of NH₃, first find the molar mass of NH₃, which is approximately 17 g/mol. Next, calculate the moles of NH₃ present in 13.6 g, and use the stoichiometry of the balanced equation to find the moles of O₂ required, which would be three-fourths of the moles of NH₃. Convert these moles of O₂ to grams using the molar mass of O₂, which is about 32 g/mol.
• b. To calculate the grams of N₂ produced from 6.50 g of O₂, first convert 6.50 g of O₂ to moles. Then use the ratio from the chemical equation to find the moles of N₂ produced, as one mole of O₂ produces two-thirds of a mole of N₂. Finally, convert these moles of N₂ to grams using its molar mass, about 28 g/mol.
• c. When determining the grams of H₂O formed from 34.0 g of NH₃, begin by calculating the moles of NH₃. Use the stoichiometry of the balanced equation, which indicates that one mole of NH₃ produces one and a half moles of H₂O. Convert these moles of H₂O to grams using the molar mass of H₂O, approximately 18 g/mol.

Answer 2

A. 19.2 g of O2.

B. 3.79 g of N2.

C. 54 g of H2O.

Step-by-step explanation:

The balanced equation for the reaction is given below:

4NH3(g) + 3O2(g) → 2N2+ 6H2O(g)

Next, we shall determine the masses of NH3 and O2 that reacted and the masses of N2 and H2O produced from the balanced equation.

This is illustrated below:

Molar mass of NH3 = 14 + (3x1) = 17 g/mol

Mass of NH3 from the balanced equation = 4 x 17 = 68 g

Molar mass of O2 = 16x2 = 32 g/mol

Mass of O2 from the balanced equation = 3 x 32 = 96 g

Molar mass of N2 = 2x14 = 28 g/mol

Mass of N2 from the balanced equation = 2 x 28 = 56 g

Molar mass of H2O = (2x1) + 16 = 18 g/mol

Mass of H2O from the balanced equation = 6 x 18 = 108 g

Summary:

From the balanced equation above,

68 g of NH3 reacted with 96 g of O2 to produce 56 g of N2 and 108 g of H2O.

A. Determination of the mass of O2 needed to react with 13.6 g of NH3.

This is illustrated below:

From the balanced equation above,

68 g of NH3 reacted with 96 g of O2.

Therefore, 13.6 g of NH3 will react with = (13.6 x 96)/68 = 19.2 g of O2.

Therefore, 19.2 g of O2 are needed for the reaction.

B. Determination of the mass of N2 produced when 6.50 g of O2 react.

This is illustrated below:

From the balanced equation above,

96 g of O2 reacted to produce 56 g of N2.

Therefore, 6.5 g of O2 will react to produce = (6.5 x 56)/96 = 3.79 g of N2.

Therefore, 3.79 g of N2 were produced from the reaction.

C. Determination of the mass of H2O formed from the reaction of 34 g of NH3.

This is illustrated below:

From the balanced equation above,

68 g of NH3 reacted to 108 g of H2O.

Therefore, 34 g of NH3 will react to produce = (34 x 108)/68 = 54 g of H2O.

Therefore, 54 g of H2O were obtained from the reaction.