Answer:
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
