Question

< Question 3 of 17 ) Strontium chloride and sodium fluoride react to form strontium fluoride and sodium chloride, according to the reaction shown. SECI, (aq) + 2 NaF(aq) SF (s) + 2 NaCl(aq) What volume of a 0.310 M NaF solution is required to react completely with 317 mL of a 0.620 M SCI, solution? volume How many moles of SrF, are formed from this reaction? moles of SrF, mol

Answer 1

syah + 2NaF lag . . " SoFz th Nad . - (aq) o we know that Hulu = 12 Miz Molarity of srlz= 0.6204 V= Volune of srile = 317 HL, ni Number of roles of ste=1 Me = Molarity of NaF= 0.310M, Vq = Volure of NaF=a ng = Vowne roles of Nafar i Ve = Mivi na si 0.620X31749 S My ni 0.310x1 hoht 1268 ML :: Voure of NaF = 1268 ML

000 12 - Molavity (ty = weight & Polet Volunettu) weight of Srila a Molarity (1) x Holdt X Volune 1000 - Org 50-6%0x158.52x317 loco = 31:157948 Moleg of sole = weight .Lt - 31.157 158.52 (trole = looo Metroles) = 0o 1965 Moles fon 196-5 Motoles weight Holeg of NaF= Molarity & Hilst x volume ..gr8 1000 = 0.310 x 46.988 X 1268 gris 1000 = 16.5 grs of NF = weight howt = 16.5948 46988 70-1989 Moles (ob 392.9Ho Roces the two Reactants srala has le88 Nuriber of sruly is 12/12 ting Reagent. anong roles. thus Cr9

among the two Reactants sraig has less Noriber of holes, thus svelg is the liniting Reagent. amount of STF?: inole of srcią (158.53 gris) produces thole of SrEz (125.6gh) 0.1965 roles of Srug ( 31.157913) Producey ? roles of $15, ca 0:1965 x1 0.1965 roles of SiFe Moles of STF2 = 0.1965 roles