For the following reaction, 9.86 grams of nitrogen gas are allowed to react with 6.06 grams of oxygen gas . nitrogen(g) + oxygen(g) nitrogen monoxide(g) What is the maximum mass of nitrogen monoxide that can be formed? grams What is the FORMULA for the limiting reagent? What mass of the excess reagent remains after the reaction is complete? grams Submit AnswerRetry Entire Group

11.34 grams of nitrogen monoxide can be formed.

The limiting reactant is O2. N2 is in excess. There will remain 4.56 grams N2.

Explanation:

Step 1: Data given

Mass of nitrogen gas = 9.86 grams

Molar mass of nitrogen gas (N2) = 28.0 g/mol

Mass of oxygen gas (O2) = 6.06 grams

Molar mass O2 = 32.0 g/mol

Step 2: The balanced equation

nitrogen(g) + oxygen(g) → nitrogen monoxide(g)

N2(g) + O2(g) → 2NO(g)

Step 3: Calculate moles

Moles = mass / molar mass

Moles N2 = 9.86 grams / 28.0 g/mol

Moles N2 = 0.352 moles

Moles O2 = 6.06 grams / 32.0 g/mol

Moles O2 = 0.189 moles

Step 4: Calculate the limiting reactant

For 1 mol N2 we need 1 moles O2 to produce 2 moles NO

O2 is the limiting reactant. It will completely be consumed (0.189 moles).

N2 is in excess. There will react 0.189 moles

There will remain 0.352 - 0.189 = 0.163 moles

Step 5: Calculate mass of excess

Mass N2 = 0.163 moles * 28.0 g/mol

Mass N2 = 4.56 grams

Step 6: Calculate moles NO

For 1 mol N2 we need 1 mol O2 to produce 2 moles NO

For 0.189 moles O2 we'll have 2* 0.189 = 0.378  moles NO

Step 7: Calculate mass NO

Mass NO = moles NO * molar mass NO

Mass NO = 0.378 moles * 30.01 g/mol

Mass NO = 11.34 grams

11.34 grams of nitrogen monoxide can be formed.

The limiting reactant is O2. N2 is in excess. There will remain 4.56 grams N2.

"standard molar enthalpy:

h2o ( liquid water ) : - 285.8 kj/mole

h2o ( water vapor ) : - 241.8 kj / mole.

δhrxn = - 241. 8 - ( - 285.8 )= -241.8 + 285.8 = 44.0

answer: 44.0 kj"