Respuesta :

kenmyna
The  volume  of  0.160    m   Li2S  solution  required  to  completely  react  with  130 ml  of 0.160  CO(NO3)2  is calculated   as  below

write the  reacting  equation

Co(NO3)2 +  Li2S = 2LiNO3  +  COS

find the    moles  of CO(NO3)2  = molarity  x  volume

=  130 ml  x  0.160=20.8  moles

since the reacting moles between CO(NO3)2  to LiS  is   1:1  the  moles of LiS  is  also  20.8  moles

volume  of Lis  is  therefore =  moles of Lis/ molarity  of LiS

=  20.8/0.160 =  130 Ml

For the complete reaction of 130 ml 0.160 M  [tex]\rm CO(NO_3)_2[/tex], 130 ml of 0.16 [tex]\rm Li_2S[/tex] is required.

The reaction of [tex]\rm Li_2S[/tex] with [tex]\rm CO(NO_3)_2[/tex] will be:

[tex]\rm Li_2S\;+\;CO(NO_3)_2\;\rightarrow\;2\;LiNO_3\;+\;COS[/tex]

Accordingly, 1 mole of [tex]\rm Li_2S[/tex] completely reacts with 1 mole of  [tex]\rm CO(NO_3)_2[/tex] .

Moles of  [tex]\rm CO(NO_3)_2[/tex]  = Molarity [tex]\times[/tex] Volume (L)

Moles of  [tex]\rm CO(NO_3)_2[/tex]  = 0.160 M  [tex]\times[/tex] 0.13 L

Moles of  [tex]\rm CO(NO_3)_2[/tex] = 0.0208 moles.

The moles of [tex]\rm Li_2S[/tex] reacts with  [tex]\rm CO(NO_3)_2[/tex]  will be 0.0208 moles.

Volume of [tex]\rm Li_2S[/tex] = [tex]\rm \frac{moles}{molarity}[/tex]

Volume of [tex]\rm Li_2S[/tex] = [tex]\rm \frac{0.0208}{0.160}[/tex]

Volume of [tex]\rm Li_2S[/tex] = 0.13 L

Volume of [tex]\rm Li_2S[/tex] = 130 ml.

For the complete reaction of 130 ml 0.160 M  [tex]\rm CO(NO_3)_2[/tex], 130 ml of 0.16 [tex]\rm Li_2S[/tex] is required.

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