4) NaCl (20.0 g) and AgNO3 (20.0 g) are dissolved in water and allowed to react (Reaction 2). I just need help with parts (d-f). PLease help me, question shown in photoInclude phase labels where appropriate and balance all equations. I also included the work I've already done for context!(Molar masses (grams/mole): Na = 23.0; Cl = 35.5; Ag = 107.9; N = 14.0; O = 16.0) ????(??) + ????ଷ (??) → ? + ? R 4) NacI (20.0 g) and AgNO3 (20.0 g) are dissolved in water and allowed to react (Reaction 2). Answerquestions (a – f). Include phase labels where appropriate and balance all equations.(Molar masses (grams/mole): Na = 23.0; Cl = 35.5; Ag = 107.9; N = 14.0; O = 16.0)%3D%3|Nacl(aq) + AgNO3(aq) → ? + ?Reaction 2Na- 23.0Write the complete molecular equation and identify the precipitate. For this you will needto predict the products of the reaction. Include phase labels and balance the equation.а.(1: 35.5Ay= 107.9N = 14.00 = 16.0Complete Ionic. Na* (aq) * CF caq) * NO caq)EquationNaci caq) +Ag NO, Caq)AgCI (s) + Na NO, (aq)%3Db. Write the complete ionic equation and identify the spectator ionsAg CICS) * Na" Caq) + NO, Caq)Spectator Ions: Nat cag) and NO3 (aq)Write the net ionic equation. (1)С.* Cross Out spect atorçi" caq) + Ag* caq) → Ag* cap * Ca" cag) * I61 caq) + Ag`CaqAg*CapAgClcslons on both sidesd. Identify and calculate the theoretical yield (grams) of the precipitate. (Hint: identify thelimiting reactant first.)AgCl cs) =precipitateIf the percent yield of the reaction is 75.0%, what mass of precipitate (grams) was formed?е.

It is given that NaCl and AgNO3 are reacting to form AgCl precipitate and NaNO3, the reaction is shown below, NaCl(aq) + AgNO3(aq) → AgCl(s) + NaNO3(aq) This is a precipitation reaction and is a type of double displacement reaction in which a solid product is formed which is insoluble in aqueous solutions,d) To determine the limiting reactant, To calculate the moles of reactant present,no. of moles = given massmolar massmoles of NaCl = 20 g58.44 g/mol = 0.342 molesmoles of AgNO3 = 20 g169.87 g/mol = 0.117 molesIn the balanced reaction as 1 moles of NaCl is reacting with 1 mole of AgNO3,Therefore,0.342 moles of NaCl will react with 0.342 moles of AgNO3,but we have 0.117 moles of AgNO3,Thus, AgNO3 is the limiting reactant,To calculate the theoretical yield of the precipitate formed, Now, 1 mole of AgNO3 is reacting to produce 1 mole of AgCl,Therefore 0.117 moles of AgNO3 will produce 0.117 moles of AgCl,Theoretical yield,= no. of moles × molar mass= 0.117 moles × 143.32 g/mol= 16.76 g e) To calculate the actual yield of AgCl when the percentage yield is 75%, percentage yield = actual yieldtheoretical yield ×100given percentage yield = 75%Therefore,75 % = actual yield16.76 g×100actual yield = 12.57 g of AgCl(precipitate) The part f is not provided in the question.