how many grams of solid KOH is needed to neutralize 1.00 L of 0.250 M HNO3 solution?
To determine the number of grams of solid KOH needed to neutralize the HNO3 solution, we can follow these steps:
Step 1: Write the balanced chemical equation for the neutralization reaction between KOH and HNO3:
KOH + HNO3 → KNO3 + H2O
Step 2: Calculate the number of moles of HNO3 in the solution.
Moles of HNO3 = Volume of solution (in liters) × Molarity of HNO3
Moles of HNO3 = 1.00 L × 0.250 mol/L = 0.250 mol
Step 3: Since the balanced equation has a 1:1 mole ratio between KOH and HNO3, the number of moles of KOH required is also 0.250 mol.
Step 4: Determine the molar mass of KOH.
Molar mass of KOH = (1 × atomic mass of K) + (1 × atomic mass of O) + (1 × atomic mass of H)
Molar mass of KOH = (1 × 39.10 g/mol) + (1 × 16.00 g/mol) + (1 × 1.01 g/mol) = 56.11 g/mol
Step 5: Calculate the mass of KOH required.
Mass of KOH = Moles of KOH × Molar mass of KOH
Mass of KOH = 0.250 mol × 56.11 g/mol = 14.03 g
Therefore, 14.03 grams of solid KOH is needed to neutralize 1.00 L of 0.250 M HNO3 solution.
Write the equation and balance it.
How many moles HNO3 do you have? That is M x L = ??
How many moles KOH are needed? From the equation you know it is the number as moles HNO3.
Then moles KOH = grams KOH/molar mass KOH. Solve for grams.