How many liters of 0.305 M K3PO4 solution are necessary to completely react with mL of 0.0184 M NiCl₂ according to the balanced chemical reaction: 187 2 K3PO4(aq) + 3 NiCl₂(aq) → Ni3(PO4)2(S) + 6 KCI(aq)

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Chapter1: Chemical Foundations
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STARTING AMOUNT
+
X
How many liters of 0.305 M K₂PO4 solution are necessary to completely react with
187 mL of 0.0184 M NiCl₂ according to the balanced chemical reaction:
mL NiCl₂
2 K3PO4(aq) + 3 NiCl₂(aq) → Ni3(PO4)2(S) + 6 KCI(aq)
ADD FACTOR
x( )
6.022 x 10²3
1
M NiCl₂
0.305
2
0.00376
3
0.00752
0.0113
ANSWER
0.001
mol NiCl₂ M K3PO4 mL K₂PO4
6
187
1000
L K3PO4
RESET
5
7.52
0.0184
g NiCl₂
g K₂PO4
Transcribed Image Text:STARTING AMOUNT + X How many liters of 0.305 M K₂PO4 solution are necessary to completely react with 187 mL of 0.0184 M NiCl₂ according to the balanced chemical reaction: mL NiCl₂ 2 K3PO4(aq) + 3 NiCl₂(aq) → Ni3(PO4)2(S) + 6 KCI(aq) ADD FACTOR x( ) 6.022 x 10²3 1 M NiCl₂ 0.305 2 0.00376 3 0.00752 0.0113 ANSWER 0.001 mol NiCl₂ M K3PO4 mL K₂PO4 6 187 1000 L K3PO4 RESET 5 7.52 0.0184 g NiCl₂ g K₂PO4
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