Question #c75aa

2 Answers
Jul 25, 2015

mass of potassium hydroxide (#KOH#)
= (1.23 x 2) (39 + 1 + 16) #g#
= 137.76 #g#


#2KOH + H_2SO_4 -> K_2SO_4 + 2H_2O#

Jul 25, 2015

Mass of potassium hydroxide needed: 138 g


Start by writing the balanced chemical equation for this neutralization reaction

#color(red)(2)KOH_((aq)) + H_2SO_(4(aq)) -> K_2SO_(4(aq)) + 2H_2O_((l))#

Notice that you have a #color(red)(2):1# mole ratio between potassium hydroxide and potassium sulfate. This means that, for every 2 moles of the former that react, the reaction will produce 1 mole of the latter.

Simply put, for a number of moles of potassium sulfate produced, twice as many moles of potassium hydroxide reacted.

You know that the reaction produced 1.23 g of potassium sulfate, which means that the number of moles of potassium hydroxide that reacted is

#1.23cancel("moles"K_2SO_4) * (color(red)(2)" moles "KOH)/(1cancel("mole"K_2SO_4)) = "2.46 moles"# #KOH#

To get the mass of potassium hydroxide that would contain this many moles, use the compound's molar mass.

A substance's molar mass tells you what the mass of 1 mole of a substance is.

#2.46cancel("moles"KOH) * "56.11 g"/(1cancel("mole"KOH)) = color(green)("138 g "KOH)#