## anonymous one year ago I need to write a net ionic equation to the following question. I have the answer below the question. How many grams of NaCl are required to precipitate most of the Ag+ ions from 2.50 × 102 mL of 0.0113 M AgNO3 solution? Write the net ionic equation for the reaction. AgNO3 moles = .0113x.250= .002825= 2.825x10-3 AgNO3=NaCl NaCl grams = 2.825x10-3x58= .1638≈.164g

Ionic equation can be written as............. $Na+ + Cl- + Ag+ + NO3- \rightarrow Ag+ + Cl- + Na+ + NO3-$ hope it'll work dude,,,,,,,,,,,,, NaCl(aq)+AgNO3(aq)→AgCl(s)+NaNO3(aq) Based on solubility rules, we know that AgCl will be (mostly) insoluble in water, so we can predict the reaction in that way. Now, we know that when NaCl and AgNO_3 dissociate, they will yield the following ions... Na+(aq)+Cl−(aq)+Ag+(aq)+NO−3(aq) To fully precipitate all of the silver(I), we need a 1:1 mole ratio of silver(I):chloride. Since we have a concentration and volume of AgNO_3, we can calculate how many moles of AgNO_3 we have... ci=niVsol′n→nAgNO3=(cAgNO3)(Vsol′n) nAgNO3=(0.0113M)(0.250L)=2.83∗10−3mol AgNO3 Each mole of silver (I) nitrate that dissociates will yield one mole of Ag^+. Based on this, we know that we have... 2.83∗10−3moles Ag+ To fully precipitate this, we need an equal number of moles of Cl^-, and therefore NaCl. Then, finally we need to convert to mass of NaCl... 2.83∗10−3mol NaCl∗58.443g NaCl1mol NaCl=0.164g NaCl