12. Balanced Chemical Equation Reaction Type: At completion of reactions: Formula of Reactant A Formula of Reactant B Formula of Product C Formula of Product D Grams of Reactant A Grams of Reactant B Grams of Product C Grams of Product D
12. Balanced Chemical Equation Reaction Type: At completion of reactions: Formula of Reactant A Formula of Reactant B Formula of Product C Formula of Product D Grams of Reactant A Grams of Reactant B Grams of Product C Grams of Product D
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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States of Matter
The substance that constitutes everything in the universe is known as matter. Matter comprises atoms which in turn are composed of electrons, protons, and neutrons. Different atoms combine together to give rise to molecules that act as a foundation for all kinds of substances. There are five states of matter based on their energies of attraction, namely solid, liquid, gases, plasma, and BEC (Bose-Einstein condensates).
Chemical Reactions and Equations
When a chemical species is transformed into another chemical species it is said to have undergone a chemical reaction. It consists of breaking existing bonds and forming new bonds by changing the position of electrons. These reactions are best explained using a chemical equation.
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Transcribed Image Text:12. Balanced Chemical Equation
Reaction Type:
At completion of reactions:
Formula of Reactant A
Formula of Reactant B
Formula of Product C
Formula of Product D
Grams of Reactant A
Grams of Reactant B
Grams of Product C
Grams of Product D
![→ Answer in FORMAT:-
2 Li + Mg(NO3)₂ → 2. LiN036
2 (2)
→ Giram of Product :- 0.61231 gm Li(NO3), and 0.00217 gm (Mg)
Single displacement Reaction
Reaction TYPE:
:-
[? of
of LiNO₂ = 13.25 mg Mg(No)₂ x 1 gm Mg(N3)2 x 1 mol Mylwes)]2
P
of mg.
= 13.25
+ Mg (This is Limming Reactant = My (NoS), is limiting Reactant)
(q)
68.94 Li NO3
Igm
-X
X
(L.R) 1x10 mg Mg(NO3)₂ 148.3 gm (Mg(NO₂)₂) 1 mul Mg(NO), 1 mul Li No₂
mg My (Nos), x 1 gm my (NO3)₂,
You will use the LIMITING REAGENT and determine
? Li Used:-
13.25 mg Mg(NO3)₂X.
Amount Remaining"
1 mol my x
1 mol My(NO3)₂
-X
X
mg
1x 10³ mg mg (NO3)₂ 148.3gm Mg(NO3)2 1 mul my (NO3), I mol
S
"Li" in Containest
= 0.01325
2 mul Li
1gm Mg(NO3)₂ 1 mol (My (NO3)2)
1x10³ mg Mg(NO3)2 148.3gm mg (NO₂)₂ ~ I mul my (143/₂" 1 mul li
X
6.94 gm Li
X
X
0.01201
2 mol LINO
gm
gm
Li
24.30gm Mg..
how much "Li"was USED:-
gm
given
0.00124
LiLEFT OVER EXCESS
Li used
= 0.01231 gm (LiNo
Theoretical Yield)
-0.00217 gm (Mg)
(Theoretical Yield)
=10-00124gm](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F74c82f9a-3d6e-4271-be5d-8c45a9bdfc0c%2Fd6760066-a4d6-47a3-98d7-bd1d4a9234c6%2Fnpg0y2_processed.jpeg&w=3840&q=75)
Transcribed Image Text:→ Answer in FORMAT:-
2 Li + Mg(NO3)₂ → 2. LiN036
2 (2)
→ Giram of Product :- 0.61231 gm Li(NO3), and 0.00217 gm (Mg)
Single displacement Reaction
Reaction TYPE:
:-
[? of
of LiNO₂ = 13.25 mg Mg(No)₂ x 1 gm Mg(N3)2 x 1 mol Mylwes)]2
P
of mg.
= 13.25
+ Mg (This is Limming Reactant = My (NoS), is limiting Reactant)
(q)
68.94 Li NO3
Igm
-X
X
(L.R) 1x10 mg Mg(NO3)₂ 148.3 gm (Mg(NO₂)₂) 1 mul Mg(NO), 1 mul Li No₂
mg My (Nos), x 1 gm my (NO3)₂,
You will use the LIMITING REAGENT and determine
? Li Used:-
13.25 mg Mg(NO3)₂X.
Amount Remaining"
1 mol my x
1 mol My(NO3)₂
-X
X
mg
1x 10³ mg mg (NO3)₂ 148.3gm Mg(NO3)2 1 mul my (NO3), I mol
S
"Li" in Containest
= 0.01325
2 mul Li
1gm Mg(NO3)₂ 1 mol (My (NO3)2)
1x10³ mg Mg(NO3)2 148.3gm mg (NO₂)₂ ~ I mul my (143/₂" 1 mul li
X
6.94 gm Li
X
X
0.01201
2 mol LINO
gm
gm
Li
24.30gm Mg..
how much "Li"was USED:-
gm
given
0.00124
LiLEFT OVER EXCESS
Li used
= 0.01231 gm (LiNo
Theoretical Yield)
-0.00217 gm (Mg)
(Theoretical Yield)
=10-00124gm
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