Question

Ammonia gas will react with oxygen gas to yield nitrogen monoxide gas and water vapor.

4NH3 + 5O2 ------> 4NO + 6H2O

a. How many moles of ammonia will react with 6.73g of oxygen?
b. If 6.42g of water is produced, how many grams of oxygen gas reacted?
c. If the reaction uses up 9.43105 g of ammonia, how many kilograms of nitrogen monoxide will be formed?
d. When 2.51 g of ammonia react with 3.76 g of oxygen, 2.27 g of water vapor are produced. What is the percentage yield of water?

Answer 1

Using the balanced chemical equation for the reaction between ammonia and oxygen, one can determine the amounts of reactants needed and products formed, as well as the percentage yield of water.

Step-by-step explanation:

The reaction between ammonia and oxygen to produce nitrogen monoxide and water is an example of a chemical reaction where stoichiometry is used to calculate the amounts of reactants and products involved. To solve the provided questions, we'll use the balanced equation 4NH3 + 5O2 → 4NO + 6H2O.

1. The number of moles of ammonia that will react with 6.73g of oxygen can be found by first converting the mass of oxygen to moles using the molar mass of O2 and then using stoichiometry to find moles of NH3.
2. If 6.42g of water is produced, the mass of oxygen gas that reacted can be calculated by converting the mass of water to moles of water, using stoichiometry to find the moles of oxygen, and then converting it back to grams.
3. Given 9.43105 g of ammonia, we can determine the mass of nitrogen monoxide formed by converting the mass of ammonia to moles, using stoichiometry to find the moles of NO, and then converting it to kilograms.
4. For a reaction with 2.51 g of ammonia and 3.76 g of oxygen where 2.27 g of water are produced, calculate the theoretical yield of water based on the reactants, and then use the actual yield (2.27 g) to find the percentage yield of water.

Answer 2

a) 0.168 moles O2

b) 9.50 grams O2

c) 0.01662 kg NO

d)88.9 %

Step-by-step explanation:

Step 1: Data given

Molar mass of NH3 = 17.03 g/mol

Molar mass of O2 = 32 g/mol

Molar mass of NO = 30.01 g/mol

Molar mass of H2O = 18.02 g/mol

Step 2: The balanced equation:

4NH3 + 5O2 → 4NO + 6H2O

a. How many moles of ammonia will react with 6.73g of oxygen?

Calculate moles of oxygen = mass O2/ molar mass O2

moles oxygen = 6.73 grams / 32.00 g/mol = 0.210 moles

Calculate moles of NH3

For 4 moles of NH3 we need 5 moles O2 to produce 4 moles NO and 6 moles H2O

For 0.210 moles O2 we need 4/5 *0.210 = 0.168 moles O2

b. If 6.42g of water is produced, how many grams of oxygen gas reacted?

Calculate moles of H2O = 6.42 grams / 18.02 g/mol = 0.356 moles

Calculate moles of O2:

For 4 moles of NH3 we need 5 moles O2 to produce 4 moles NO and 6 moles H2O

For 0.356 moles H2O we'll need 5/6 * 0.356 = 0.297 moles O2

Calculate mass of O2 = moles O2 * molar mass O2

Mass O2 = 0.297 moles O2 * 32.00 g/mol = 9.50 grams O2

c. If the reaction uses up 9.43105 g of ammonia, how many kilograms of nitrogen monoxide will be formed?

Calculate moles of ammonia = 9.43105 grams / 17.03 g/mol =0.5538 moles

Calculate moles of NO:

For 4 moles of NH3 we need 5 moles O2 to produce 4 moles NO and 6 moles H2O

For 0.5538 moles of NH3 we'll have 0.5538 moles NO

Calculate mass of NO

Mass NO = 0.5538 moles * 30.01 g/mol = 16.62 grams = 0.01662 kg NO

d. When 2.51 g of ammonia react with 3.76 g of oxygen, 2.27 g of water vapor are produced. What is the percentage yield of water?

Calculate moles of NH3 = 2.51 grams / 17.03 g/mol = 0.147 moles

Calculate moles of O2 = 3.76 grams / 32 g/mol = 0.118 moles

Determine the limiting reactant

O2 is the limiting reactant, it will completely be consumed (0.118 moles)

NH3 is in excess. There will react 4/5 * 0.118 = 0.0944 moles

There will remain 0.147 - 0.0944 = 0.0526 moles

Calculate moles H2O: For 0.118 moles O2 we'll have 6/5 * 0.118 = 0.1416 moles H2O

Calculate mass H2O = 0.1416 moles * 18.02 g/mol = 2.552 grams H2O

Calculate % yield = (2.27/2.552)*100 % = 88.9 %