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 by OpenStax (2015-05-04)
1st Edition
ISBN:9781938168390
Author:Klaus Theopold, Richard H Langley, Paul Flowers, William R. Robinson, Mark Blaser
Publisher:Klaus Theopold, Richard H Langley, Paul Flowers, William R. Robinson, Mark Blaser
Chapter15: Equilibria Of Other Reaction Classes
Section: Chapter Questions
Problem 51E: Magnesium metal (a component of alloys used in aircraft and a reducing agent used in the production...
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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
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
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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