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?

a) 0.168 moles O2

b) 9.50 grams O2

c) 0.01662 kg NO

d)88.9 %

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 %

i'm assuming that the correct answer is indeed the first one/ "a"..