How many grams of KOH are needed to neutralize 12.9 mL of 0.16 M HCl in stomach acid?

Well, that's quite a chemistry question you've got there! It seems like you're trying to bring some balance into the world of stomach acid. So, let's calculate the amount of KOH needed.

To find the grams of KOH, we'll need to use the formula:

moles = concentration x volume

First, let's convert the volume from milliliters (mL) to liters (L). We can do this by dividing 12.9 mL by 1000 to get 0.0129 L.

Now, let's use the given concentration, 0.16 M, and multiply it by the volume in liters:

moles = 0.16 M x 0.0129 L = 0.002064 moles

Since we're dealing with a 1:1 ratio of KOH to HCl when it comes to neutralization, we can say that 0.002064 moles of KOH are needed.

Next, we need to find the molar mass of KOH, which is roughly 56.11 g/mol. So, to calculate the grams of KOH needed:

grams = moles x molar mass = 0.002064 moles x 56.11 g/mol ≈ 0.116 grams

Therefore, approximately 0.116 grams of KOH are needed to neutralize 12.9 mL of 0.16 M HCl in stomach acid. But hey, remember that I'm just a clown bot, so take my answer with a pinch of humor!

To determine the number of grams of KOH needed to neutralize the HCl, we'll use the following steps:

Step 1: Write the balanced equation for the neutralization reaction:
HCl + KOH → KCl + H2O

Step 2: Calculate the number of moles of HCl:
Molarity (M) = Moles (mol) / Volume (L)
0.16 M = Moles (HCl) / 0.0129 L

Moles (HCl) = 0.16 M x 0.0129 L
Moles (HCl) = 0.002064 mol

Step 3: From the balanced equation, we can see that the stoichiometric ratio between HCl and KOH is 1:1. Therefore, the number of moles of KOH required will be the same as the moles of HCl.

Step 4: Calculate the mass of KOH:
Molar mass of KOH = 39.1 g/mol (K) + 16.0 g/mol (O) + 1.01 g/mol (H)
Molar mass of KOH = 56.11 g/mol

Mass of KOH = Moles (KOH) x Molar mass of KOH
Mass of KOH = 0.002064 mol x 56.11 g/mol
Mass of KOH = 0.1159 g

Therefore, approximately 0.1159 grams of KOH are needed to neutralize 12.9 mL of 0.16 M HCl in stomach acid.

To determine the number of grams of KOH needed to neutralize the given amount of HCl, we can follow these steps:

1. Write the balanced chemical equation for the reaction between KOH and HCl:
1 KOH + 1 HCl → 1 KCl + 1 H₂O

2. Calculate the number of moles of HCl present in the given solution using its molarity and volume:
Moles of HCl = Molarity × Volume
= 0.16 M × 12.9 mL

Note: Since the volumes are given in milliliters (mL), it's important to convert mL to liters (L) by dividing by 1000.

3. Calculate the number of moles of KOH required to neutralize the calculated moles of HCl:
According to the balanced equation, the stoichiometric ratio between HCl and KOH is 1:1. Thus, the number of moles of KOH is equal to the number of moles of HCl.

4. Determine the molar mass of KOH:
The molar mass of K (potassium) is approximately 39.10 g/mol.
The molar mass of O (oxygen) is approximately 16.00 g/mol.
The molar mass of H (hydrogen) is approximately 1.01 g/mol.
Adding these values together gives us the molar mass of KOH, which is approximately 56.11 g/mol.

5. Calculate the mass of KOH:
Mass of KOH = Moles of KOH × Molar mass of KOH

By following these steps, we can determine the mass of KOH required to neutralize the given amount of HCl.

blub?

.0129 L * 0.16 M = 1.9E-3 moles

multiply 1.9E-3 by molar mass of KOH