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)
