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Allushta [10]
3 years ago
6

What volume of 0.205 M K3PO4 solution is necessary to completely react with 134 mL of 0.0102 M NiCl2?

Chemistry
1 answer:
viktelen [127]3 years ago
5 0
 the balanced equation for the above reaction is as follows;
3NiCl₂ + 2K₃PO₄ ---> Ni₃(PO₄)₂ + 6KCl
stoichiometry of NiCl₂ to K₃PO₄ is 3:2
the number of NiCl₂ moles reacted - 0.0102 mol/L x 0.134 L = 0.00137 mol
according to molar ratio 
if 3 mol of NiCl₂ reacts with 2 mol of K₃PO₄
then 0.00137 mol of NiCl₂ reacts with - 2/3 x 0.00137 = 0.000911 mol of K₃PO₄

molarity of given K₃PO₄ solution - 0.205 M
there are 0.205 mol in 1000 mL 
therefore volume of 0.000911 mol - 0.000911 mol / 0.205 mol/L = 4.44 mL 
volume of K₃PO₄ needed is 4.44 mL
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\displaystyle \frac{n_1}{V_1} = \frac{n_2}{V_2}

Substitute in known values:

\displaystyle \frac{(3.31 \text{ mol})}{(100 \text{ L})}  = \frac{(13.15\text{ mol })}{V_2}

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2- 0.16 M/s.

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<em>The rate of the reaction = - d[PH₃]/4dt = d[P₄]/dt = d[H₂]/6dt.</em>

where, - d[PH₃]/dt is the rate of PH₃ changing "rate of disappearance of PH₃".

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