Question

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

Answer 1

There will be produced 30.64 grams of nitrogen dioxide

The statement is false

Step-by-step 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