Question

1. For the reaction... H2SO4 + 2LIOH Li2SO4 + 2H2O a. What volume of 3.98 M H2SO4 is required to neutralize 25.0 grams of LiOH? b. If 25.50 mL of 0.625 M LiOH is required to react with 35.50 mL of H2SO4, what is the molarity of the H2SO4? 2. For the following reaction... Na2SO4 + BaCl2 2 NaCl + BaSO4 a. If 125 mL of a 3.25 M BaCl2 solution reacts with Na2SO4, how many grams of BaSO4 is produced? b. What is the maximum amount in grams of BaSO4 that can be produced from 25. mL of a .125 M Na2SO4 solution and 50.0 mL of a .250 M BaCl2 solution?

Answer 1

H2SO4 + 2L10H Liz sou + 2H2O Molan mass of LiOH = 23.94 gmol Moles of Linh = 25.09 23.94 gmolt on = 1.044 mo? rioles of H₂SO4 required for neutralization - 1 mol H2SO4 x 1.044 mon lioh 2mo Li OH = 0.522 mol H2SO4 Volume of H2SO4 required = 0.522 mon 3.98 more! = 0.1312 L =(0.1312 81000) me = 1131-2 m2)

b) Moen g li OH = 25.0ml x 0.625 m = 25.50 (x 0.625 mole 1600 = 0.01594 mo H₂SO4 required : 0.01594 mon notes of 0.225 m = 0.00797 mon molarity g HSO4 = 0.000 m 35.50mL 0.00797 a (35.50. 1000 2. Na2SO4 + Bach - Znac + Basou - Noles of Ball = 0.125 LX 3.25 m/ = 0.40625 mon As Back and Basou are in Ill moon ratio, moles q Basou formed = 0.40625 mm Mass q BaSO4 formed = 0.40625 mod molar mass = 0.40625 m07x 233.3 qmo' = 194.89 b) Mores of Na S0u = 125.0 J L x 0.125 mol = 0.003/25 inn 1600/

Moles of BaSO4 produced by Na2SO4 = 0.003125 mor (Nas 50, and Basou) are lil mora ratio) roles of Back = 1500) x 0.250 mo/ = 0.0725 mon So, moles q BaSO4 produced from Back = 0.0125 mm The roles of BaSO4 produced is lesser in Case q Na2SO4. So, Na2SO4 res the limiting reactant. Now maximum amorent of Baso 4 that can be produced a 0.003125 m07x 233.3 9/2007 = 10.729 g