Respuesta :
Answer:
0.538 M of Cl- in solution
Explanation:
First we need the balanced reaction:
2AgNO3 + CaCl2 ⇒ Ca(NO3)2 + 2AgCl
Then , we calculate mol of Ag in the reaction.
mol Ag= 450mL * (0.242 molAgNO3/ 1000mL) = 0.1089 mol AgNO3
Now, how many moles of CaCl2 we need to react with 0.1089 mol AgNO3? We take this information, from balanced reaction.
0.1089 mol AgNO3* (1 mol CaCl2/ 2 mol AgNO3) = 0.05445 mol CaCl2
We just need 0.054 mol CaCl2, and the rest of CaCl2 remain in solution.
so,
0.08 mol CaCl2 - 0.05445 mol CaCl2 = 0.0256 mol CaCl2 in excess,
Now, how many moles of Cl- are in 0.0256 mol of CaCl2?
0.0256 mol CaCl2* 2 mol Cl-/1mol CaCl2= 0.0511 mol Cl-
Molarity Cl- = Moles Cl-/ Volumen (L)
V (L) = (450 mL + 400 mL) * (1L/ 1000 mL) = 0.950 L
M = 0.0511mol Cl-/ 0.950 L = 0.538 M
Chlorides help compensate around 0.05 percent of the earth's mantle. Freshwater usually contains chloride concentrations range from 1 - 100 ppm (parts per million). Ionic compounds are soluble in water in subterranean aquifers, which are geological structures that hold groundwater, and further calculation can be defined as follows:
Calculating the [tex]\bold{AgNO_3}[/tex] volume:
[tex]\to \bold{V = 450.0 \ mL = 0.45\ L}\\\\[/tex]
Calculating the number of mol in [tex]\bold{AgNO_3}[/tex]:
[tex]\to \bold{n = Molarity \times Volume}[/tex]
[tex]= 0.242 \times 0.45\\\\= 0.1089\ mol[/tex]
Calculating the [tex]\bold{[CaCl_2]}[/tex] volume:
[tex]\to \bold{V = 400.0 \ mL= 0.4 \ L}\\\\[/tex]
Calculating the number of mol in [tex]\bold{[CaCl_2]}[/tex]:
[tex]\to \bold{n = Molarity \times Volume}[/tex]
[tex]= 0.2\times 0.4\\\\= 8 \times 10^{-2}\ mol[/tex]
Balancing equation:
[tex]2 AgNO_3 + CaCl_2 \longrightarrow 2 AgCl + Ca(NO_3)_2[/tex]
When 2 mol of [tex]\bold{AgNO_3}[/tex] reacts with 1 mol of [tex]\bold{[CaCl_2]}[/tex] for 0.1089 mol of [tex]\bold{AgNO_3}[/tex] [tex]6.638 \times 10^{-2}[/tex] mol of [tex]\bold{[CaCl_2]}[/tex] is required But we have [tex]8 \times 10^{-2}[/tex]mol of [tex]\bold{[CaCl_2]}[/tex] so, [tex]\bold{AgNO_3}[/tex] is limiting reagent From balanced chemical reaction, we see that when 2 mol of [tex]\bold{AgNO_3}[/tex] reacts, 1 mol of [tex]\bold{[CaCl_2]}[/tex] reacts.
Calculating reacted mol of [tex]\bold{[CaCl_2]}[/tex] [tex]= (\frac{1}{2})\times AgNO_3\ mole[/tex]
[tex]= (\frac{1}{2})\times 0.1089\\\\=0.05445\\\\= 5.445\times 10^{-2}\ mol[/tex]
Calculating remaining mol of [tex]\bold{[CaCl_2]}[/tex] = [tex]\text{initially presented mol - reacted mol }[/tex]
[tex]= 8\times 10^{-2} - 5.445 \times 10^{-2} \\\\= 2.555 \times 10^{-2} mol[/tex]
Calculating the total volume:
[tex]\bold{= 450.0\ mL + 400.0 \ mL}\\\\ \bold{= 850.0\ mL}[/tex]
Calculating the remaining [tex]\bold{[CaCl_2]}[/tex] [tex]=\frac{ \text{remaining mol of}\ CaCl_2}{ volume}[/tex]
[tex]=\frac{2.555 \times 10^{-2}\ mol}{ 0.850\ L}\\\\= 0.0217175\ M[/tex]
Calculating the remaining [tex]\bold{[Cl^{-}] }[/tex] :
[tex]\bold{= 2 \times 0.0217175\ M}\\\\ \bold{= 0.043435\ M}[/tex]
Learn more:
brainly.com/question/6917019
