Question

Solutions of sodium carbonate and silver nitrate reacts to form silver carbonate and sodium nitrate. A solution containing 2.35 g of sodium carbonate is mixed with one containing 12.8 g of silver nitrate. Show your work. Pay close attention to significant figures and units. Use 4 significant figures for the masses. Answers without work will receive a grade of 0 pts a) Write the chemical equation 2.359 Nalo 2 (sodem corrono b) Balanced the molecular equation c) Write the net ionic equation 2.09 AaNO3 (swr nitrato d) Determine the limiting reactant e) Calculate the amount of silver carbonate formed. a) Na2CO₃ + AgNO₃ →

Answer 1

Chemical Equation I daCO₃ + AG NO 3 This equation is based - NaNO3 + AgaCO3 on your question. 6 Balanced Molecolar Equation. first we write we balance from Reaction Chemical equation and then each atom in the molecule on both the sides, das (03 (aq) +2Ag No₃ lag) 2daNO3(aq) + Ag (03/5) the (O3 iors Ag toote No3 involved = 1 = 1x2 = 1x2 = =2 = 2 Naz 1x2 = CO3 = 1 Ag a 2 = do3 = I x3 = 2 To balance da we multiply it by a n Rts, so the 'No3 is also got multplied by and on Lts we multiply Ag by 2 and NO3 got multiplied by a so. LHS = RMS, and the equation is balanced

det lonEic reaction first we will write complete ionic reaction f with the help of Babaced moleculax equate Bellanced Molecular equation is das 103 (aq) + 2 Ag Noz lag) - 2 dadoz 199) + (7 636). complete ionic Equation is 2 dat (aq) + (0 (99) + 2 Ag tagst 2 No3 (ag) 2 Nat caq) + 2dog (aq) + Aga (03 (s) AgaCO3 is a solid is so it is written as it is. and to write net ionic equation we will elaminate spectator ions. So The det ionic equation is 2 Ag+ (aq) + (03² (aq) Ag2 103 (s) d) To determine limiting Reagent, we will need to find out moler of Reactants. moles of das 103 = Amount (in grams) Molecular weight of Molar Mass = 2.35 g (given 105.ag 0.0222 mole

Moles of AgNO3 = Arount lingrams) - Molecular weight or rolar mass = 12-8 (g) 169. 89 = 0.0753 moles Na2 (03 (aq) + 2AgNO3(aq) dado (aq) + Ag (03 (6) So by the stoichiomexy we can say that I moles of daCO3 laq) required 2 moles of AgNO3. and 0.0222 moles of dan (03 requires = 0.0222x2 = 0.0444 neles o AgNO3 and we have 0-0753 moles of Agd3 so the liesiting reagent is Na₂03 because AgNO3 is in excess amount Ans :- Naa (O3 is a limiting Reagent. Amount of silver Carbonate formed we can solve this by stoichiomery agarr And we will pick Na2CO3 because it is a limiting Reagent! Na C03 (aa) + 2Ag Nozlaa) 2dado 3 (aq) + Ag 10318) here, we know that moles of Naz (os forms ! mole d Ag CO₂. So, 0-0 222 moles of Na2CO3 will form 0.0222 moles of AgCO3. The amount of Ag 103 fooms = 0.0222 * Molecular Mars = 0.0222 x 284.69 = 6-3 18 grams