1.) Aluminum Metal, Al, reacts with oxygen gas, O 2 , to produce aluminum oxide,
ID: 1008344 • Letter: 1
Question
1.) Aluminum Metal, Al, reacts with oxygen gas, O2, to produce aluminum oxide, Al2O3. Which reactant is the limiting reagent when 35.0 grams of Al reacts with 35.0 grams of O2? (Use AL2O3 as your reference product.)
4Al (s) + 3O2 (g) ---> AL2O3 (s)
2.) Nitrogen dioxide gas, NO2, reacts with water to produce nitric acid, HNO3, and nitrogen monoxide, NO. How many grams of nitrogen monoxide are produced when 100 grams of Nitrogen dioxide gas reacts with 100 grams of water? (Use limiting reagent)
3NO2 (g) + H2O (l) ---> 2HNO3 (aq) + NO (g)
Explanation / Answer
(1) 4Al (s) + 3O2 (g) ---> Al2O3 (s)
Molar mass of Al is 27 g/mol
Molar mass of O2 is = 2xAt.mass of O = 2x16 = 32 g/mol
According to the balanced reaction ,
4 moles of Al reacts with 3 moles of O2
OR
4x27=108 g of Al reacts with 3x32=96 g of O2
35.0 g of Al reacts with M g of O2
M = ( 35.0x96) / 108
= 31.1 g of O2
So 35.0 - 31.1 = 3.9 g of O2 left unreacted so it is the excess reactant.
Since all the mass of Al completly reacted it is the limiting reactant.
(2) 3NO2 (g) + H2O (l) ---> 2HNO3 (aq) + NO (g)
Molar mass of NO2 = At.mass of N + ( 2xAt.mass of O)
= 14 + ( 2x16)
= 46 g/mol
Molar mass of H2O = ( 2xAt.mass of H) + At.mass of O
= ( 2x1) + 16
= 18 g/mol
Molar mass of NO = At.masss of N + At.mass of O
= 14 + 16
= 30 g/mol
According to the balanced reaction,
3 moles=3x46 g of NO2 reacts with 1 mol=1x18 g of water
100 g of NO2 reacts with M g of water
M = ( 100x18) / (3x46)
= 13.04 g of water
So 100-13.04 = 86.96 g of water left unreacted so it is the excess reactant.
Since all the mass of NO2 completly reacted it is the limiting reactant.
From the balanced reaction ,
3 moles of NO2 produces 1 mole of NO
OR
3x46 g of NO2 produces 1x30 g of NO
100 g of NO2 produces N g of NO
N = ( 100x30) / ( 3x46)
= 21.7 g of NO
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.