If this reaction is conducted with 57.4 g CoCl₂ and 42.0 g F₂, how many grams of I mol Cla= 129.8 g/mol) = 38.0 g/mol g/mol .ماما = 2 1) 57.4 & coclax. = 0.4422 mol = 1.1053 mo 0.4422 mol Cocta x Imol Cofa Imol Colta 42.09 F2 x 1 mol 38.09 0= 134.7 g/mol =3 = 71.0 g/mol 03= 76.01 g/mol 129.88 1.1053 mol x 1 mol CoF₂ = 1.1053 mol Cof₂ I mol from Fa 29.29 Cofa 0.4422 mol Coffa from CoCl2 0.4422 mol Cofax 66.09 Imol 2. Balance this equation: 3 Sno+ 2 NF₁ → 3 SnF2 + _N₂0₂ If this reaction is performed using 102.3 g SnO and 75.6 g NF3, how many grams of N₂O, will produced? 1) 102.3 g SnOx Imol = 0.7595 134.79 = 1.0648 I mel N₂O3 75.6g NF3 NF3 x Imol 71.09 react 2532 mol N₂O3 limiting reactant

Chemistry & Chemical Reactivity
10th Edition
ISBN:9781337399074
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Chapter3: Chemical Reactions
Section3.2: Balancing Chemical Equations
Problem 3.1CYU: (a) Butane gas, C4H10, can burn completely in air [use O2(g) as the other reactant] to give carbon...
icon
Related questions
Question

Could someone check these two stoichiometry problems and tell me if I did them correctly? Thank you so much for your feedback!!

1. Balance this equation:
COCI+FCF₂+ Ch
If this reaction is conducted with 57.4 g CoCl₂ and 42.0 g F₂, how many grams of CoF₂ will be produced?
1) 57.4 g Coclax.
I mol
= 0.4422 mol CoCla
= 1.1053 mol F₂
COCI2 129.8 g/mol)
F₂= 38.0 g/mol
COF2= 66.0 g/mol)
(2) 0.4422 mol Cocta x
1.1053 mol x 1 mol Cof₂
I mol
·=
3) 0.4422 mol Cota x
Sno= 134.7 g/mol
NF3 =
N203
=
42.09 Fax 1 mol
38.09
Imol Cofa
Imol Colta
71.0 g/mol
76.01 g/mol)
2) 0.7595
mot no X
1.0648
mol NF3 X
66.09
Imol
2. Balance this equation: 3 Sno+ 2 NF₁ → 3 SnF₂+_ __N₂03
=
I mol N₂O3
3 mot Sno
0.4422 mol Coffa
from CoCl2
If this reaction is performed using 102.3 g SnO and 75.6 g NF₁, how many grams of N₂O, will be
produced?
Imol
SnO x
1) 102.3 g
I mol N₂03
2 mot NF3
129.88
=
75.69 NF3
1.1053 mol COF₂
from Fac
29.29 Cofa
X
=
134.79
Imol
71.09
=
= 0.2532 mol N₂O3 limiting
from Sno
reactant
= 0.5324 mol N₂03
from NF3
reactant
0.7595 mol Sno
1.0648 mol NF3
3) 0.2532 mol N₂03 x 76.019 19.25 g N₂03
I mol
Scanned with CamScanner
Transcribed Image Text:1. Balance this equation: COCI+FCF₂+ Ch If this reaction is conducted with 57.4 g CoCl₂ and 42.0 g F₂, how many grams of CoF₂ will be produced? 1) 57.4 g Coclax. I mol = 0.4422 mol CoCla = 1.1053 mol F₂ COCI2 129.8 g/mol) F₂= 38.0 g/mol COF2= 66.0 g/mol) (2) 0.4422 mol Cocta x 1.1053 mol x 1 mol Cof₂ I mol ·= 3) 0.4422 mol Cota x Sno= 134.7 g/mol NF3 = N203 = 42.09 Fax 1 mol 38.09 Imol Cofa Imol Colta 71.0 g/mol 76.01 g/mol) 2) 0.7595 mot no X 1.0648 mol NF3 X 66.09 Imol 2. Balance this equation: 3 Sno+ 2 NF₁ → 3 SnF₂+_ __N₂03 = I mol N₂O3 3 mot Sno 0.4422 mol Coffa from CoCl2 If this reaction is performed using 102.3 g SnO and 75.6 g NF₁, how many grams of N₂O, will be produced? Imol SnO x 1) 102.3 g I mol N₂03 2 mot NF3 129.88 = 75.69 NF3 1.1053 mol COF₂ from Fac 29.29 Cofa X = 134.79 Imol 71.09 = = 0.2532 mol N₂O3 limiting from Sno reactant = 0.5324 mol N₂03 from NF3 reactant 0.7595 mol Sno 1.0648 mol NF3 3) 0.2532 mol N₂03 x 76.019 19.25 g N₂03 I mol Scanned with CamScanner
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 3 steps with 2 images

Blurred answer
Knowledge Booster
Combustion Analysis
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Similar questions
Recommended textbooks for you
Chemistry & Chemical Reactivity
Chemistry & Chemical Reactivity
Chemistry
ISBN:
9781337399074
Author:
John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:
Cengage Learning
Chemistry & Chemical Reactivity
Chemistry & Chemical Reactivity
Chemistry
ISBN:
9781133949640
Author:
John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:
Cengage Learning
Chemistry: An Atoms First Approach
Chemistry: An Atoms First Approach
Chemistry
ISBN:
9781305079243
Author:
Steven S. Zumdahl, Susan A. Zumdahl
Publisher:
Cengage Learning
Chemistry: The Molecular Science
Chemistry: The Molecular Science
Chemistry
ISBN:
9781285199047
Author:
John W. Moore, Conrad L. Stanitski
Publisher:
Cengage Learning
Chemistry
Chemistry
Chemistry
ISBN:
9781305957404
Author:
Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:
Cengage Learning
Chemistry
Chemistry
Chemistry
ISBN:
9781133611097
Author:
Steven S. Zumdahl
Publisher:
Cengage Learning