how many grams of baking soda is needed to neutralize 500ml of hydrochloric acid
?
Depends upon the strength of the HCl doesn't it?
NaHCO3 + HCl ==> H2O + CO2 + NaCl
To determine the amount of baking soda needed to neutralize hydrochloric acid, you need to consider the balanced chemical equation for the reaction between baking soda (sodium bicarbonate, NaHCO3) and hydrochloric acid (HCl):
NaHCO3 + HCl -> NaCl + H2O + CO2
From the equation, we can see that one mole of baking soda reacts with one mole of hydrochloric acid to produce one mole of carbon dioxide, one mole of water, and one mole of sodium chloride.
Step 1: Calculate the moles of hydrochloric acid:
To find the moles of hydrochloric acid, we need to know the concentration of the acid. Let's assume it is a 1 M (1 mole per liter) solution.
Using the formula:
Moles = Concentration (M) * Volume (L)
Moles = 1 M * 0.5 L (since 500 ml equals 0.5 L)
Moles = 0.5 mol
Therefore, we have 0.5 moles of hydrochloric acid.
Step 2: Calculate the moles of baking soda:
From the balanced equation, we can see that the stoichiometric ratio between baking soda and hydrochloric acid is 1:1. Therefore, the number of moles of baking soda required will be the same as the moles of hydrochloric acid.
Moles of baking soda = 0.5 mol
Step 3: Calculate the mass of baking soda:
To find the mass of baking soda, we need to know the molar mass of baking soda, which is 84.01 g/mol.
So,
Mass of baking soda = Moles of baking soda * Molar mass of baking soda
Mass of baking soda = 0.5 mol * 84.01 g/mol
Mass of baking soda = 42.005 g
Therefore, you would need approximately 42 grams of baking soda to neutralize 500 ml of hydrochloric acid.