Nitric acid,HNO3, is available ata a concentration of 16 M. how much would react with 5.58 g of KOH acording to the following equation?

HNO3+KOH-H2O+KNO3

dont understand? does not give me a volume amount to work with and I really don't understand how to solve it

I responded to this in your earlier post except I see I used 5.56 g and not 5.58 g.

http://www.jiskha.com/display.cgi?id=1290380148

To solve this problem, you will need to use stoichiometry to determine the amount of nitric acid (HNO3) that reacts with the given amount of KOH.

1. Determine the molar mass of KOH:
- The molar mass of K (Potassium) is 39.10 g/mol.
- The molar mass of O (Oxygen) is 16.00 g/mol.
- The molar mass of H (Hydrogen) is 1.01 g/mol.
- Therefore, the molar mass of KOH is 39.10 + 16.00 + 1.01 = 56.11 g/mol.

2. Convert the mass of KOH to moles:
- Divide the given mass of KOH (5.58 g) by its molar mass (56.11 g/mol):
5.58 g KOH / 56.11 g/mol = 0.0993 mol KOH

3. Determine the molar ratio between HNO3 and KOH using the balanced equation:
- From the balanced equation, 1 mole of HNO3 reacts with 1 mole of KOH.

4. Calculate the moles of HNO3 that can react:
- Since the molar ratio is 1:1, the moles of HNO3 that can react will be the same as the moles of KOH: 0.0993 mol.

5. Determine the volume of 16 M HNO3 that contains 0.0993 moles:
- Use the equation: Molarity (M) = moles / volume (in liters)
- Rearrange the equation to solve for volume:
Volume = moles / Molarity
- Plug in the values:
Volume = 0.0993 mol / 16 M = 0.00621 L or 6.21 mL

Therefore, approximately 6.21 milliliters of 16 M nitric acid (HNO3) would react with 5.58 grams of KOH according to the given equation.