Can someone please explain this problem to me step by step pleaseeeeeee?

In the following acid—base neutralization reaction

HNO3 (aq)+ KOH (aq)---KNO3(aq) + H2O(I)

What is the molarity (M) of an HNO3 solution if 50.0ml is needed to react with 25.0ml of 0.150M KOH solution?

How many mols KOH do you have? That's mols KOH = M x L = ?

How many mols HNO3 will that neutralize. The same number since the equation is 1 mol HNO3 to 1 mol KOH.

M HNO3 = mols HNO3/L HNO3.

To find the molarity (M) of the HNO3 solution, we can use the concept of stoichiometry. Here's the step-by-step solution:

Step 1: Write the balanced equation for the reaction.
HNO3 (aq) + KOH (aq) → KNO3 (aq) + H2O (l)

Step 2: Determine the molarity of the KOH solution.
Given: Volume of KOH solution = 25.0 mL, Molarity of KOH = 0.150 M

Step 3: Use the volume and molarity of KOH to calculate the number of moles of KOH.
Moles of KOH = Volume (L) x Molarity (mol/L)
Moles of KOH = (25.0 mL / 1000 mL/L) x 0.150 mol/L

Step 4: Use the balanced equation to determine the mole ratio between KOH and HNO3.
From the balanced equation:
1 mole of KOH reacts with 1 mole of HNO3

Step 5: Convert the moles of KOH to moles of HNO3 using the mole ratio.
Moles of HNO3 = Moles of KOH

Step 6: Use the volume and moles of HNO3 to calculate the molarity of the HNO3 solution.
Molarity of HNO3 = Moles of HNO3 / Volume of HNO3 (L)
Volume of HNO3 is given as 50.0 mL.

Once you complete these steps, you will have the molarity (M) of the HNO3 solution.