Question

Consider the equation 2KOH + H2SO4 → K2SO4 + 2H20 a lf 25 g H2SO4 is reacted with 7.7 g KOH, how many grams of K2SO4 are produced? b For part a of this problem, identify the limiting reactant and calculate the mass of excess reactant that remains after the reaction is completed. c Calculate the theoretical yield of the reaction. How many grams of material would you expect to obtain if the reaction has a 67.1% yield?

Answer 1

a) Anewer = 11.95 grams Explanation given data mass of H₂SO4 = 25g mals of KOH = 7.7 g Standard values required are molar mass of H₂SO4 = 98.07 g. mot Molar mass of KOH= 56.10 g.mol' molarmars of K₂SO4 = 174-26 gemolo Balanced equation g KOH + H₂SO - K₂SO4 + 2H0 - 2 mole Imole i mole To find out the grams of K₂.604 produced, we need to know the limiting reagent, as explained below mass ..7.7g. moles of koH = Molar Mass in = 0.1372 mole 56.10.ginal 258 moles of H₂SO4 = 0 98.07 g moll = 0.2549 mole tan moles of H₂SO4 since moles of Koh is less than moles of t KOH is limiting reagent - moles of KOH_0.1372 mole moles of K₂SO4 = moles of K2504 = 0.0686 mole

Mass of K₂SO4 = Moles et K₂SO4* Molar mass of Kasa en = 0.0686 pol x 174.26 gomfort mass at Kaso4 = 11.95 g [theoretical field] b) as we explained in part-a, Limiting reactant is the one which has less number of moles 1. so koH is limiting reactant mass of excess reactant H₂SO remained = 11.54 g Explanation as per the balanced equation, only 0.1372 moles of Koh reacts with 0.1372 moles of H₂SO, Hence 0.1177 moles of H₂SO4 will remain so mass of 0.1177 mole H₂SO4 can be calculated as below mole = mass Molar mats 0.1177 wote = - 98.07 g.mol-1 x = 0.1177 mofle x 98.07 ginoplet x= 11.54 g

c) theoretical yield is 11.95 g, we alleadly calculated in part-a calculation, This is the maximum output of given reaction if the limiting reactant is 100% converted into product (the dretically) given that 67.1% yield; so this can be calculated as below 100% Yield = 11.95 g 67.1% yield = x 11.95 g x 67.1 x - x= 100% x = 8.018 g we get . If the reaction has a 67.1% yield - 8.018 g of producto