In neutralization reactions,
acid + base �¨ salt + water
HCl can be neutralized with NaOH. How many moles of NaOH will it take to neutralize 0.68 L of 0.29 M HCl?
Is there a formula that I need to use here?
HCl + NaOH --- > H2O + NaCl balanced as is
One mole of NaOH for every one mole of HCl because the equation balances with one of each
.68 * .29 = .1972 moles of HCl so we need .1972 moles of NaOH
To calculate the number of moles of NaOH needed to neutralize a given amount of HCl, you can use the formula:
Moles = Volume (in liters) × Molarity
Here's how you can apply this formula to solve the given problem:
Step 1: Identify the given values
- Volume of HCl: 0.68 L
- Molarity of HCl: 0.29 M
Step 2: Calculate the moles of HCl
Moles of HCl = Volume × Molarity
= 0.68 L × 0.29 M
= 0.1972 moles HCl
Step 3: Remember that in a neutralization reaction, the ratio between the moles of acid and base will be 1:1. Therefore, the moles of NaOH needed will be the same as the moles of HCl.
So, the number of moles of NaOH needed to neutralize 0.68 L of 0.29 M HCl is 0.1972 moles.