Respuesta :
Answer:
1) 2.0 g
2) 0 g
3) 4.17 g
4) 2.57 g
Explanation:
First of all, we need to know the compounds and the reaction. The ion carbonate is [tex]CO3^{-2}[/tex], and the ion nitrate is [tex]NO3^{-}[/tex].
Sodium is in group 1, so it must lose one electron to be stable, and be the cation [tex]Na^{+}[/tex]. Silver has only one electron too, so the cation will be [tex]Ag^{+}[/tex].
To form the chemical compounds, first we put the cation, then the anion, and change their charges without the signal:
Sodium carbonate: Na2CO3
Silver nitrate: AgNO3
Silver carbonate: Ag2CO3
Sodium nitrate: NaNO3
The balanced reaction will be:
Na2CO3 + 2 AgNO3 --> Ag2CO3 + 2 NaNO3
Now, we must check the stoichiometry, which will be 1:2:1:2 (always in number of moles)
The question wants to know the mass value, so we need to know the molar mass of these compounds. Checking the periodic table will see that:
Na = 23 g/mol, C = 12 g/mol, N = 14 g/mol, O = 16 g/mol, Ag = 108 g/mol
So the molar mass of the compounds must be:
Na2CO3 = 106 g/mol (2x23 + 12 + 3x16)
AgNO3 = 170 g/mol (108 + 14 + 3x16)
Ag2CO3 = 276 g/mol (2x108 + 12 + 3x16)
NaNO3 = 85 g/mol
We have a mixture of the reactants, so one probably would be in excess, so, first will need to test. Let's do the stoichiometry calculus using silver nitrate as the limit, so:
1 mol of Na2CO3 ---------- 2 mol of AgNO3
106 g ------------------------------ 2x170 = 340 g
x ------------------------------------ 5.14 g
By a simple direct three rule:
340x = 544.84
x = 1.6 g of Na2CO3
That means that for this reaction, we only need 1.6 g of Na2CO3 to react with 5.14 of AgNO3. How we have 3.60 g of Na2CO3, it is on excess, and all the AgNO3 will be consumed.
1) The mass of Na2CO3 that remains after the reaction will be the initial less the mass that reacted:
m = 3.6 - 1. 6 = 2.0 g
2) All the AgNO3 reacted, so there isn't a mass present after the reaction.
m = 0 g
3) Now, doing the stoichiometry calculus between AgNO3 and Ag2CO3
2 moles of AgNO3 ------------- 1 mol of Ag2CO3
2x170 g ------------------------------- 276 g
5.14 g --------------------------------- x
By a simple direct three rule:
340x = 1418.64
x = 4.17 g of Ag2CO3
4) Now, doing the stoichiometry calculus between AgNO3 and NaNO3
2 moles of AgNO3 ----------------------- 2 moles of NaNO3
2x170 g ---------------------------------------- 2x85 g
5.14 g ------------------------------------------- x
By a simple direct three rule:
340x = 873.8
x = 2.57 g
2.0 grams of sodium carbonate, 0 gram of silver nitrate, 4.17 grams of silver carbonate and 2.57 grams of sodium nitrate is present after the reaction is complete.
How we calculate no. of grams from moles?
We can calculate Number of grams or wanted weight through below expression:
[tex]\[{\rm{n = }}\frac{{\rm{W}}}{{\rm{M}}}\][/tex], where
n = no. of mole
W = given weight
M = molar mass
In the question given balanced chemical reaction is:
[tex]${\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{C}}{{\rm{O}}_{\rm{3}}}{\rm{ + 2AgN}}{{\rm{O}}_{\rm{3}}} \to {\rm{A}}{{\rm{g}}_{\rm{2}}}{\rm{C}}{{\rm{O}}_{\rm{3}}}{\rm{ + 2NaN}}{{\rm{O}}_{\rm{3}}}$[/tex]
It is clear that mole ratio of reactant and product is 1:2::1:2.
Also mention that, given initial mass of sodium carbonate = 3.60g
And given initial mass of silver nitrate = 5.14g
In the reactant mixture one would be present in the excess quantity and we did stoichiometry calculation by assuming silver nitrate as the limiting reactant, so:
1 mole of Na₂CO₃ ............................. 2 mole of AgNO₃
106 g .............................. 2 × 170 = 340 g
x .............................. 5.14 g
Now by using simple three rule, we can calculate the value of x:
340x = 544.84
x = 1.6 g of Na₂CO₃
(1). Mass of Na₂CO₃ remains after the completion of reaction:
Initial - used = 3.6 - 1.6 = 2.0 g
(2). All the AgNO₃ get consumed by the excess Na₂CO₃, so mass present after completion of reaction is 0 g.
(3). Now again we will do stoichiometry calculation between AgNO₃ and Ag₂CO₃:
2 moles of AgNO₃ ............................ 1 mole of Ag₂CO₃
2 × 170 g ............................. 276 g
5.14 g ............................. x
Now by using simple three rule, we can calculate the value of x:
340x = 1418.64
x = 4.17 g of Ag₂CO₃
(4). Again we will do stoichiometry calculation between AgNO₃ and NaNO₃:
2 moles of AgNO₃ .............................. 2 moles of NaNO₃
2 × 170 g .............................. 2 × 85 g
5.14 g ............................... x
Now by using simple three rule, we can calculate the value of x:
340x = 873.8
x = 2.57 g
Hence, amount of sodium carbonate, silver nitrate, silver carbonate and sodium nitrate is 2.0 g, 0 g, 4.17 g and 2.57 g respectively.
To learn more about moles, visit this below link:
https://brainly.com/question/1034638
