20.0 g of Nitrogen is produced when oxygen gas reacts with NO gas. 29.8 L of oxygen is required at STP to produce this 20.0 g of NO. A. yes B. false

There will be produced 30.64 grams of nitrogen dioxide

The statement is false

Explanation:

Step 1: Data given

MAss of nitrogen produced = 20.0 grams

Molar mass of N2 = 28.0 g/mol

Mass of NO = 20.0 grams

Molar mass of NO = 30.01 g/mol

Volume of O2 at STP = 29.8 L

Step 2: the balanced equation

2NO + O2 → 2N02

Step 3: Calculate moles NO

Moles NO = mass NO / molar mas NO

Moles NO = 20.0 grams / 30.01 g/mol

Moles NO = 0.666 moles

Step 4: Calculate moles O2

1 mol O2 at STP = 22.4 L

29.8 L = 29.8/22.4 = 1.33 moles

Step 5: Calculate the limiting reactant

For 2 moles NO we need 1 mol O2 to produce 2 moles NO2

NO is the limiting reactant. IT will completely be consumed (0.666 moles).

O2 is in excess. There will react 0.666/2 = 0.333 moles O2

There will remain 1.333 - 0.333 = 0.997 moles O2

Step 6: Calculate moles NO2

For 2 moles NO we need 1 mol O2 to produce 2 moles NO2

For 0.666 moles NO we'll have 0.666 moles NO2

Step 7: Calculate mass NO2

Mass NO2 = moles NO2 * molar mass NO2

Mass NO2 = 0.666 moles * 46.0 g/mol

Mass NO2 = 30.64 grams

There will be produced 30.64 grams of nitrogen dioxide

The statement is false

the enthalpy of formation of water is given as -285.8kj/mol, which states that the enthalpy of formation is negative.

negative enthalpy of formation indicates that the heat is released during the reaction process. the heat is released from reaction to the surroundings the enthalpy is expressed with a negative sign with the energy given out in the reaction process.

the formation of water takes place by giving out heat which is -285.8 kj/mol.