For the following acid-base reaction, calculate the mass (in grams) of each acid necessary to completely react with and neutralize 5.00 g of the base.
HCl(aq)+ NaOH(aq)->H20(l)+NaCl(aq)
Please explain how to do this.
To calculate the mass of each acid necessary to completely react with and neutralize 5.00 g of the base, we need to use stoichiometry. Stoichiometry is the calculation of the quantitative relationship between reactants and products in a chemical reaction.
First, we need to determine the mole ratio between the base (NaOH) and the acid (HCl) in the balanced chemical equation. The balanced equation shows that 1 mole of NaOH reacts with 1 mole of HCl.
Next, we calculate the molar mass of the base (NaOH) and the acid (HCl) to convert the given mass of the base (5.00 g) to moles.
The molar mass of NaOH is:
Na: 22.99 g/mol
O: 16.00 g/mol
H: 1.01 g/mol
So, the molar mass of NaOH is 22.99 + 16.00 + 1.01 = 40.00 g/mol.
To calculate the number of moles of NaOH, we divide the given mass by the molar mass:
Number of moles of NaOH = 5.00 g / 40.00 g/mol = 0.125 mol.
Since the mole ratio between NaOH and HCl is 1:1, we know that 0.125 moles of NaOH will react with 0.125 moles of HCl.
Finally, we convert the moles of HCl to grams using the molar mass of HCl, which is 36.46 g/mol:
Mass of HCl = 0.125 mol × 36.46 g/mol = 4.56 g.
Therefore, you would need approximately 4.56 grams of HCl to completely react with and neutralize 5.00 grams of NaOH in this acid-base reaction.
Here is a worked example of a stoichiometry problem. Just follow the steps.
http://www.jiskha.com/science/chemistry/stoichiometry.html