A nitric acid solution is neutralized using sodium hydroxide. How many grams of sodium hydroxide are needed to neutralize 5.50 L of 1.200 M nitric acid solution?

HNO3 (aq) + NaOH (aq) → H2O (l) + NaNO3 (aq)

How many moles HNO3 do you have? M x L = moles.

Using the coefficients in the balanced equation, convert moles HNO3 to moles KOH.

Then moles KOH = grams KOH/molar mass KOH; solve for grams KOH.

To find out how many grams of sodium hydroxide (NaOH) are needed to neutralize the nitric acid solution, you need to follow a series of steps:

Step 1: Write and balance the chemical equation.
The balanced chemical equation for the neutralization reaction between nitric acid (HNO3) and sodium hydroxide (NaOH) is:
HNO3 (aq) + NaOH (aq) → H2O (l) + NaNO3 (aq)

Step 2: Convert the given volume of the nitric acid solution into moles.
Given: Volume of nitric acid solution = 5.50 L
Concentration of nitric acid (HNO3) = 1.200 M

Moles of HNO3 = Concentration (M) × Volume (L)
= 1.200 mol/L × 5.50 L
= 6.60 moles

Step 3: Equate the moles of HNO3 to the moles of NaOH using the balanced equation.
From the balanced equation, we can see that 1 mole of HNO3 reacts with 1 mole of NaOH.
Therefore, the number of moles of NaOH required is also 6.60 moles.

Step 4: Convert moles of NaOH into grams.
To convert moles of NaOH to grams, we need to know the molar mass of NaOH, which is the sum of the atomic masses of sodium (Na), oxygen (O), and hydrogen (H).
The atomic masses are: Na = 22.99 g/mol, O = 16.00 g/mol, H = 1.01 g/mol.

Molar mass of NaOH = Na (22.99 g/mol) + O (16.00 g/mol) + H (1.01 g/mol)
= 39.00 g/mol + 16.00 g/mol + 1.01 g/mol
= 56.01 g/mol

Now, we can find the mass of NaOH using the equation:
Mass (g) = Moles × Molar mass
= 6.60 moles × 56.01 g/mol
= 369.96 grams

Therefore, 369.96 grams of sodium hydroxide (NaOH) are needed to neutralize 5.50 L of 1.200 M nitric acid solution.