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What volume of 0.181 M #Na_3PO_4# solution is necessary to completely react with 91.0 mL of 0.107 M #CuCl_2#?

Answer 1

Ellie Andrews

#35.9"ml"#

Let's begin with the equation that is balanced:

#3CuCl_(2(aq))+2Na_3PO_(4(aq))rarrCu_3(PO_4)_(2(s))+6NaCl(aq)#

This tells us that 3 moles of #CuCl_(2)# react with 2 moles #Na_3PO_4#.

#:.# 1 mole #CuCl_2# will react with 2/3 moles #Na_3PO_4#

Concentration equals moles divided by volume, as we know.

#c=n/v#

#:.n=cxxv#

#:. nCuCl_2=0.107xx91.0/1000=9.737xx10^(-3)#

I divided by 1000 to convert #"ml"# to #"L"#

#:.nNa_3PO_4=9.737xx10^(-3)xx2/3=6.491xx10^(-3)#

#v=n/c=(6.491xx10^(-3))/(0.181)=35.86xx10^(-3)"L"#

#:.v=35.86"ml"#

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Answer 2

Xavier Avery

To find the volume of 0.181 M Na3PO4 solution required to react with 91.0 mL of 0.107 M CuCl2, you need to use the stoichiometry of the reaction between Na3PO4 and CuCl2.

The balanced chemical equation for the reaction is:

[ 2 \text{ Na}_3\text{PO}_4(aq) + 3 \text{ CuCl}_2(aq) \rightarrow \text{ Cu}_3(\text{PO}_4)_2(s) + 6 \text{ NaCl}(aq) ]

Use the stoichiometry of the reaction to find the moles of CuCl2 and then determine the volume of Na3PO4 solution needed based on its concentration.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.