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

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