What mass of KOH is required to neutralize 200mL of 3.16 M HNO3?
find the mole for HNO3 using n = MV where M is the molarity and V is the volume in Liters.
then use the balance equation;
KOH + HNO3 --> KNO3 + H2O
the reaction is 1:1 so the mole you calculated for HNO3 is also the mole for KOH.
i.e. mole KOH = mole HNO3.
then convert the mole to mass using the molar mass of KOH (obtain from periodic table) to get the mass for KOH.
hope that helps...
To find the mass of KOH required to neutralize the given amount of HNO3, we need to first determine the stoichiometry of the reaction.
The balanced equation for the reaction between KOH (potassium hydroxide) and HNO3 (nitric acid) is:
KOH + HNO3 -> KNO3 + H2O
From the equation, we can see that 1 mole of KOH reacts with 1 mole of HNO3 in a 1:1 ratio.
Since we know the concentration and volume of the HNO3 solution, we can calculate the number of moles of HNO3 present using the formula:
moles = concentration (M) x volume (L)
First, convert the volume of the HNO3 solution from milliliters (mL) to liters (L):
200 mL = 200/1000 = 0.2 L
Now we can calculate the number of moles of HNO3:
moles of HNO3 = 3.16 M x 0.2 L = 0.632 moles
Since we have a 1:1 stoichiometric ratio between KOH and HNO3, the same number of moles of KOH is required to neutralize the HNO3.
Finally, to find the mass of KOH needed, we need to use the molar mass of KOH. The molar mass of KOH is:
K (39.1 g/mol) + O (16.0 g/mol) + H (1.0 g/mol) = 56.1 g/mol
Now we can calculate the mass of KOH:
mass of KOH = moles of KOH x molar mass of KOH
mass of KOH = 0.632 moles x 56.1 g/mol
mass of KOH ≈ 35.5 grams
Therefore, approximately 35.5 grams of KOH is required to neutralize 200 mL of 3.16 M HNO3.