How many milliliters of .100 M HNO3 are needed to neutralize 58.5 mL of .0100 Al(0H)3?

17.6 mL

3HNO3 + Al(OH)3 ==> Al(NO3)3 + 3H2O

How many mols Al(OH)3 do you have? That is M x L = ?

So how many mols HNO3 are needed to neutralize it. The equations tells you moles HNO3 = 3* moles Al(OH)3.

Then M HNO3 = moles HNO3/L HNO3.
You have M and moles solve for L HNO3 an convert to mL.

Well, it seems like we have a reaction between nitric acid (HNO3) and aluminum hydroxide (Al(OH)3). Let's calculate the amount of HNO3 needed to neutralize Al(OH)3.

First, we need to find the number of moles of Al(OH)3:
58.5 mL of 0.0100 M Al(OH)3 is equal to (0.0585 L) x (0.0100 mol/L) = 0.000585 mol Al(OH)3.

Since the balanced equation for this reaction is:
3 HNO3 + Al(OH)3 → Al(NO3)3 + 3 H2O

We can see that it takes 3 moles of HNO3 to react with 1 mole of Al(OH)3. So, we need 3 times the moles of Al(OH)3 in HNO3.

Therefore, the number of moles of HNO3 needed is:
0.000585 mol Al(OH)3 x 3 mol HNO3/mol Al(OH)3 = 0.001755 mol HNO3.

Now, let's find the volume of the 0.100 M HNO3 solution needed to contain 0.001755 mol.
Volume (in L) = moles / concentration
Volume = 0.001755 mol / 0.100 mol/L = 0.01755 L.

Finally, we can convert this volume to milliliters:
Volume = 0.01755 L x (1000 mL/1 L) = 17.55 mL.

So, approximately 17.55 milliliters of 0.100 M HNO3 are needed to neutralize 58.5 mL of 0.0100 M Al(OH)3.

To determine the volume of .100 M HNO3 needed to neutralize 58.5 mL of .0100 M Al(OH)3, we first need to write a balanced equation between the acid and base:

Al(OH)3 + 3HNO3 → Al(NO3)3 + 3H2O

From the balanced equation, we can see that 1 mole of Al(OH)3 reacts with 3 moles of HNO3.

Next, we need to calculate the number of moles of Al(OH)3 in 58.5 mL of .0100 M Al(OH)3.

Moles of Al(OH)3 = volume (in liters) × molarity
= 0.0585 L × 0.0100 mol/L
= 0.000585 mol

Since 1 mole of Al(OH)3 reacts with 3 moles of HNO3, we can calculate the number of moles of HNO3 needed:

Moles of HNO3 = 3 × moles of Al(OH)3
= 3 × 0.000585 mol
= 0.001755 mol

Now, we can calculate the volume of .100 M HNO3 needed using the moles of HNO3 and the given molarity:

Volume of HNO3 = moles ÷ molarity
= 0.001755 mol ÷ 0.100 mol/L
= 0.01755 L

Finally, we convert the volume from liters to milliliters:

Volume of HNO3 = 0.01755 L × 1000 mL/L
= 17.55 mL

Therefore, approximately 17.55 mL of .100 M HNO3 is needed to neutralize 58.5 mL of .0100 M Al(OH)3.

To determine the number of milliliters of 0.100 M HNO3 needed to neutralize 58.5 mL of 0.0100 Al(OH)3, we need to use the concept of stoichiometry and the balanced chemical equation.

The balanced chemical equation for the reaction between HNO3 and Al(OH)3 is:

3 HNO3 + Al(OH)3 -> Al(NO3)3 + 3 H2O

From the balanced equation, we can see that it takes 3 moles of HNO3 to react with 1 mole of Al(OH)3. Therefore, the stoichiometric ratio between HNO3 and Al(OH)3 is 3:1.

Now, let's solve the problem step-by-step:

1. Convert the given volumes to liters:
- 58.5 mL of Al(OH)3 = 58.5 mL / 1000 mL/L = 0.0585 L

2. Calculate the number of moles of Al(OH)3:
- moles of Al(OH)3 = volume in liters * molarity of Al(OH)3
= 0.0585 L * 0.0100 mol/L = 0.000585 mol

3. Use the stoichiometric ratio to determine the number of moles of HNO3 needed:
- moles of HNO3 = moles of Al(OH)3 * (3 moles HNO3 / 1 mole Al(OH)3)
= 0.000585 mol * (3/1) = 0.00176 mol

4. Convert the moles of HNO3 to milliliters using the molarity of HNO3:
- volume of HNO3 = moles of HNO3 / molarity of HNO3
= 0.00176 mol / 0.100 mol/L
= 0.0176 L * 1000 mL/L
= 17.6 mL

Therefore, it requires 17.6 mL of 0.100 M HNO3 to neutralize 58.5 mL of 0.0100 Al(OH)3.