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

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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
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