Question

LAB VII. PRACTICE PROBLEMS - CHAPTER 4 NAME: SECTION: b) How many grams of strontium nitrate, Sr(NO), are needed to prepare 3.40x10 ml. of a 0.925 M solution? c) To what volume should 50.0mL of a 6.00M HCI stock solution be diluted to obtain a 1.50M HCl solution? How much water should be used? d) A 38.5ml sample of 0.875M AgNO, solution is mixed with 51.0mL sample of 0.744M NaCl solution. The AgCl precipitated has a mass of 2.06g. Calculate the limiting reactan the theoretical yield, and percent yield. AgNO3(aq) + NaCl) → AgCl + NaNO30) III) Types of Aqueous Solutions and Solubility a) Circle the compound(s) AgCl, CaSO or NaC,H,O, that is (are) insoluble in water b) Circle the strong electrolyte H2SO, or CH 06- c) List the ions present in a soluble solution of sodium sulfate. IV) Complete and balance the following equations: a) (gas evolution) HCI + Na SO, → b) (acid-base) H2SO4 + KOH → c) (precipitation) Sr(OH)2 Fe(NO3) = V) Write the ionic and net ionic equation for equations b) and c) of exercise IV. b) lonic: Net ionic: Ionic: Net ionic:

Answer 1

v = volume of Srino, Me molarity of ST (NO) solnion = 3.40 x 103 m solution 2 0.925M VE 3.40 x10 mn b = 3.40 xro3x10-31 = 3.40 L (: Im L= 303 mL) na moles of Ss(NO3)2 = MXV = 0.925 M X 3.40 L In= 3.145 moles) We know that, molar mass of Sr(NO3/2 = 212 g/nd i mass of Sr(NO3)= moles of Sr (NO₂), molar mass of 1 Sr(NO₃)2 mass of Srino), = 3.145 mal x 21291 mal mans of SolNo? = 666.749 & 1667g] M = initial conch of nel= 6.00M M = final conen of HCl = 1.50M v = volume of not initially so.om N2= Volume of hel finally = ? we know that 1. IM xv=constant lino. of moles remains . M, V, M V , Constant) V2 - 6.00 MX 50.0 mL - 1.50M = 200.mL CS Scanned with CamScanner

volume of ware added final volume Initial blume of her 04 HU volume of water = 200.mL - 50.om! .. volume of water a 150.mL Given deaction, AgNO3(aq) + nacl coq) — Agu (s) + NONO, CaqJ Given Memolarity of AgNO₃ = 0.875M. Ni = volume of AG No₂ = 38.5 mL M₂ = molarity of Noll= 0.7449 ₂ = volume of Nach = 51.0mL n, moles o ANO, = M,XV = 0.875MX38.5 mL .n33.7 millimoley h₂= moles of Nach in M₂ X V₂ = 0.744M X slomi 0.74MX 51.0 mL n2 = 37.9 millimoles. we senow that, fruitting reactant is the reactant which chas one least value of moles aovided by stochiometric roefficient for AG NO3, - moles Stochometry - 3304 = 33.7 For Nach moles = Stochometry 37.9 = 37.9 Scanned with CamScanner

we see least value of moles divedel by stoctions is for AgNo, hence, AgNO is limiting reacant By s ro chiomering of reachion, I mover of AgNo gives i mole of Agu ... 33.7 milfmoles o Agnez gines 33.9 millimeles Aga. :, moles ol ng 11 = 33.7 X10-3 mol/ we know that, molau Man Agus 143. mass of Agei= moles of ageix molar mas of ngel. mass of Agcl = 33.7x 103 mal x 143.5g/mol mass of age = 4.3%,9 :. Theorential yield is 14.849]. Given, mass of Angel = 2.06g. . Achcel yield is 2.009]. we know that % yield: Actual gield The orchical yield . . ytelde 2.06 X100 4.84 % yield: 0.426 X 100 = 142.6% CS Scanned with CamScanner Scan
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