Calculate the grams of KHP needed to react with 25.2 ml of 0.10M KOH if the reaction is?
KOH+KHP -----> K2P + H2O
moles KOH = M x L
Use the equation to convert moles KOH to moles KHP.
moles KHP = grams/molar mass. Solve for grams.
To calculate the grams of KHP (potassium hydrogen phthalate) needed to react with 25.2 ml of 0.10M KOH (potassium hydroxide), you need to use the balanced equation and perform a stoichiometric calculation.
The balanced equation you provided is:
KOH + KHP → K2P + H2O
The molar ratio between KHP and KOH in the balanced equation is 1:1, which means that 1 mole of KHP reacts with 1 mole of KOH.
First, we need to calculate the number of moles of KOH using the given volume and concentration:
moles of KOH = volume (in liters) × concentration (in moles per liter)
moles of KOH = 0.0252 L × 0.10 mol/L
moles of KOH = 0.00252 moles
Since the mole ratio between KHP and KOH is 1:1, the number of moles of KHP required is also 0.00252 moles.
Next, we can calculate the mass of KHP using its molar mass. The molar mass of KHP is 204.23 g/mol.
mass of KHP = moles of KHP × molar mass of KHP
mass of KHP = 0.00252 mol × 204.23 g/mol
mass of KHP = 0.5117 grams
Therefore, approximately 0.5117 grams of KHP are needed to react with 25.2 ml of 0.10M KOH in the given reaction.