Respuesta :
Answer:- (a) [tex]O_2[/tex] , (b) 1.73 g NO and 1.55 g [tex]H_2O[/tex] , (c) 0.932 g of [tex]NH_3[/tex] , (d) yes, the results are consistent with law of conservation of mass.
Solution:- The given balanced equation is:
[tex]4NH_3+5O_2\rightarrow 4NO+6H_2O[/tex]
From balanced equation, ammonia and oxygen react in 4:5 mole ratio means 4 moles of ammonia reacts with 5 moles of oxygen to give the products.
(a) Let's calculate the moles of the reactants on dividing the grams by molar masses as:
For ammonia, [tex]1.90gNH_3(\frac{1mole}{17g})[/tex] = 0.112 mole
For oxygen, [tex]2.30gO_2(\frac{1mole}{32g})[/tex] = 0.0719 mole
From mole ratio, we need more moles of oxygen as compared to oxygen but from calculations the moles of oxygen are less than ammonia. So, oxygen is limiting reactant.
(b) To calculate the grams of each product formed we would calculate their moles first from limiting reactant that is oxygen and then multiply by their respective molar masses to convert the moles to grams as:
For grams of NO:
[tex]0.0719moleO_2(\frac{4moleNO}{5moleO_2})(\frac{30gNO}{1moleNO})[/tex] = 1.73 g of NO
For grams of [tex]H_2O[/tex] :
[tex]0.0719moleO_2(\frac{6moleH_2O}{5moleO_2})(\frac{18gH_2O}{1moleH_2O})[/tex] = 1.55 g of [tex]H_2O[/tex]
So, 1.73 grams of NO and 1.55 grams of [tex]H_2O[/tex] would formed.
(c) Let's calculate the moles of excess reactant that is ammonia used to react with limiting reactant, oxygen as:
[tex]0.0719moleO_2(\frac{4moleNH_3}{5moleO_2})[/tex]
= 0.0572 mole [tex]NH_3[/tex]
Excess moles of ammonia = 0.112 - 0.0572 = 0.0548 mole
Now let's convert these excess moles of ammonia to grams on multiplying by molar mass:
[tex]0.0548moleNH_3(\frac{17g}{1mole})[/tex]
= 0.932 g [tex]NH_3[/tex]
So, after reaction when all the oxygen is used then 0.932 grams of ammonia would still be remaining.
(d) mass of reactants = mass of products + remaining mass of excess reactant
Let's plug in the values in it:
mass of reactants = 1.90 g + 2.30 g = 4.2 g
mass of products + remaining mass of excess reactant = 1.73 g + 1.55 g + 0.932 g
= 4.2 g
So, the results are consistent with law of conservation of mass.
The limiting reactant is this reaction is oxygen.
The equation of the reaction is;
4NH3(g) + 5O2(g) -----> 4NO(g) + 6H2O(g)
Number of moles of NH3 = 1.90 g/17 g/mol = 0.11 moles
Number of moles of O2 =2.30 g/32 g/mol = 0.07 moles
To find the limiting reactant;
4 moles of NH3 reacts with 5 moles of O2
0.11 moles of NH3 reacts with 0.11 moles × 5 moles/4 moles
= 0.14 moles
We can see that there is less O2 than required hence O2 is the limiting reactant.
If 5 mols of O2 yield 4 moles of NO
0.07 moles of O2 yield 0.07 moles × 4/5
= 0.056 moles of NO
Mass of NO formed = 0.056 moles × 30 g/mol
= 1.73 g
If 5 moles of O2 yield 6 moles of H2O
0.07 moles of O2 yield 0.07 moles × 6 moles/5 moles
= 0.084 moles
Mass of water formed = 0.084 moles × 18 g/mol = 1.55 g
4 moles of NH3 reacts with 5 moles of O2
x moles of NH3 reacts with 0.07 moles of O2
x = 0.056 moles
Number of moles of excess reactant = 0.11 moles - 0.056 moles
=0.054 moles
Mass of excess reactant left= 0.054 moles × 17 g/mol = 0.92 g
According to the law of conservation of mass, total mass before reaction must be equal to total mass after reaction.
Total mass before reaction= 1.90 g + 2.30 g = 4.2 g
Total mass after reaction = 0.92 g + 1.73 g + 1.55 g = 4.2 g
Hence the law of conservation of mass has been demonstrated.
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