The equation for this precipitation reaction using the formulas of the reactant*

Question

The equation for this precipitation reaction using the formulas of the reactant* and products is 2AgNO_3(aq) + CaCI_2(aq) 2AgCI(a) + Ca(NO_3)(aq) Solid AgCl forms according to the following net ionic equation: Ag- (aq) + CP (aq) AgCl(s) The number of moles of these ions initial solutions are: Because Ag and CI react in a 1:1 ratio, the amount of Ag" will be limiting (0.0380 mole Ag* is less than 0.0920 mole of CI). Since Ag" is limiting, only 0.0380 mole of solid AgCl will be formed. Therefore, the mass of AgCl can be calculated: Calculate the concentrations of each ion remaining in solution after precipitation is complete. (If no ions remain, leave the box blank and click on Submit.) Concentration of Ag+ M Concentration of CI^- M Concentration of Ca2+ W Concentration of NO_3 M

Explanation / Answer

After reaction competition concentration o f [Ag+] =0

1 mole CaCl2 = 2 mole Cl-

Moles of Cl- = 0.200 L * 2* 0.23 Moles/L

= 0.0920 moles

Now calculate the moles of Cl- which are remain in the recation mixture=

0.0920 moles - 0.0380 moles = 0.054 moles Cl-

Total volume = 200 .0 ml+200.0 ml

= 400.0 ml

= 0.400 L

Molarity = number of moles / volume in L

= 0.054 moles Cl- / 0.400    L

= 0.135 M

NO3- was not reacted it is present as NO3-

Moles of NO3-   = 0.200 L * 0.19 Moles/L

= 0.0380 moles

Molarity = number of moles / volume in L

= 0.0380 Moles / 0.400 L

= 0.0950 M

Ca2 + was not reacted it is present as Ca 2+

Moles of Ca2+ = 0.200 L * 0.23 Moles/L

= 0.0460 moles

Molarity = number of moles / volume in L

= 0.046 Moles / 0.400 L

= 0.115 M

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