#color (green) "Step 1:"#
The first thing you need to do is write the chemical equation.
#KOH# + #H_2SO_4# = #K_2SO_4# + #H_2O# (unbalanced)
#color (green) "Step 2:"#
Balance the chemical equation. You need to do this because it's the only way that you can get the "conversion factor" you needed to solve the problem.
#color (red) 2 KOH# + #H_2SO_4# = #K_2SO_4# + #color (red) 2H_2O# (balanced)
left side:
K = 1 x #color (red) 2# = 2
O = (1 x #color (red) 2#) + 4 = 6
H = (1 x #color (red) 2#) + 2 =4
S = 1
right side:
K = 2
O = 4 + (1 x #color (red) 2#) = 2
H = 2 x #color (red) 2# = 4
S = 1
#color (green) "Step 3:"#
Convert the volume of #H_2SO_4# into number of moles. How? By multiplying the given volume to its given molarity. Notice that I already converted the unit from milliliter (mL) to liter (L).
volume #H_2SO_4# = 500 mL = 0.5 L
concentration of #H_2SO_4# = 2M or #2 mol*L^"-1"#
#0.5 cancel L# #H_2SO_4# x #"2 moles"/(1 cancel "L") ##H_2SO_4# = 1 mole #H_2SO_4#
#color (green) "Step 4:"#
Now, going back to the balanced equation,
#color (red) 2 KOH# + #color (red) 1 H_2SO_4# = #color (red) 1 K_2SO_4# + #color (red) 2H_2O#
we know that
#color (red) 1# mole #H_2SO_4# = #color (red) 2# moles #KOH#
Therefore, to convert the computed mole of #H_2SO_4# into mole of #KOH#,
#1# #cancel "mol"##H_2SO_4# x #(2 "mol" KOH)/(1 cancel ("mol") H_2SO_4)# = #2# #"mol"# #KOH#
#color (green) "Step 5:"#
Getting the atomic weights from the periodic table, we can compute the molar mass of #KOH#,
K = #39.10 "g/mol"#
O = #16.00 "g/mol"#
H = #1.01 "g/mol"#
molar mass of #KOH# = #39.10 "g/mol"# + #16.00 "g/mol"# + #1.01 "g/mol"# = #56.11 "g/mol"#
#color (green) "Step 6:"#
Converting the number of moles we got from #color (green) "Step 4"# using the molar mass of #KOH# obtained in #color (green) "Step 5"#,
#2# #cancel "mol"##KOH# x #(56.11 grams KOH)/(1 cancel ("mol") KOH)# = #color (red) (112.22 g KOH)#
