can you help me plug it in the right form i keep doing it and its wrong ↑60.0escapp.101edu.comL AgNO, XSTARTING AMOUNT60.0 mL AgNO, xX960.500 mol AgCl1 mol AgCl0F13.662X-0³F20.500A1143.32 g AgNO₂1 mol AgCl#mol AgClmol AgCl80F3How many grams of AgCl will be formed when 60.0 mL of 0.500 M AgNO iscompletely reacted according to the balanced chemical reaction:153 mol AgCl3 mol AgClNOV7$FeCl₂(aq) +g AgClXADD FACTOR6.022 x 1023x( )F4= 4.30 g AgCl4.30143.3210.001%L AgNO33169.880F5XAgNO₂(aq)- 3 AgCl(s) + Fe(NO₂)₂(aq)g AgNO,Question 3 of 4mol AgClDELETEH143.32LAgClάtv1—0.500F6X3=M AgNO,mol AgNO, g AgNO,ANSWER&MacBook Air1.43383F71000mol AgClmol AgCl mL AgNO₂mol AgCl60.04.h ZA12.93RESETDIIF821.70mL AgCl= 4.30*8F9(((g AgCl30F10-WF11&+0+F12

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Answered: How many grams of AgCl will be formed…

How many grams of AgCl will be formed when 60.0 mL of 0.500 M AgNO, is completely reacted according to the balanced chemical reaction: FeCl₂(aq) + 3 AgNO,(aq) 3 AgCl(s) + Fe(NO₂),(aq)

Introductory Chemistry: A Foundation

9th Edition

ISBN:9781337399425

Author:Steven S. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Donald J. DeCoste

Chapter15: Solutions

Section: Chapter Questions

Problem 34QAP: 34. For each of the following solutions, the number of moles of solute is given, followed by the...

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can you help me plug it in the right form i keep doing it and its wrong

↑
60.0
esc
app.101edu.co
mL AgNO, X
STARTING AMOUNT
60.0 mL AgNO, x
X
96
0.500 mol AgCl
1 mol AgCl
0
F1
3.662
X
-0³
F2
0.500
A
1
143.32 g AgNO₂
1 mol AgCl
#
mol AgCl
mol AgCl
80
F3
How many grams of AgCl will be formed when 60.0 mL of 0.500 M AgNO is
completely reacted according to the balanced chemical reaction:
15
3 mol AgCl
3 mol AgCl
NOV
7
$
FeCl₂(aq) +
g AgCl
X
ADD FACTOR
6.022 x 1023
x( )
F4
= 4.30 g AgCl
4.30
143.32
1
0.001
%
L AgNO3
3
169.88
0
F5
X
AgNO₂(aq)- 3 AgCl(s) + Fe(NO₂)₂(aq)
g AgNO,
Question 3 of 4
mol AgCl
DELETE
H
143.32
LAgCl
άtv
1
—
0.500
F6
X
3
=
M AgNO,
mol AgNO, g AgNO,
ANSWER
&
MacBook Air
1.43
3
8
3
F7
1000
mol AgCl
mol AgCl mL AgNO₂
mol AgCl
60.0
4.h ZA
12.9
3
RESET
DII
F8
2
1.70
mL AgCl
= 4.30
*
8
F9
(((
g AgCl
30
F10
-
W
F11
&
+
0
+
F12

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