What mass of hydrochloric acid (in grams) can be neutralized by 2.7g of sodium bicarbonate?
Here is a simple stoichiometry problem I've posted. Just follow the steps.
http://www.jiskha.com/science/chemistry/stoichiometry.html
To find the mass of hydrochloric acid that can be neutralized by sodium bicarbonate, we need to use the concept of stoichiometry.
Stoichiometry is the relationship between the relative quantities of substances involved in a chemical reaction. In this case, the balanced chemical equation for the reaction between sodium bicarbonate (NaHCO3) and hydrochloric acid (HCl) is:
2 NaHCO3 + HCl -> NaCl + CO2 + H2O
From the balanced equation, we can see that 2 moles of sodium bicarbonate react with 1 mole of hydrochloric acid. To find the mass of HCl neutralized, we need to convert the given mass of sodium bicarbonate (2.7g) into moles first.
The molar mass of sodium bicarbonate (NaHCO3) is:
Na: 22.99 g/mol
H: 1.01 g/mol
C: 12.01 g/mol
O: 16.00 g/mol
3 O atoms in HCO3: 3 * 16.00 = 48.00 g/mol
Adding all the molar masses together:
NaHCO3 = 22.99 + 1.01 + 12.01 + 48.00 = 84.01 g/mol
Now we can calculate the number of moles of sodium bicarbonate:
moles of NaHCO3 = mass of NaHCO3 / molar mass of NaHCO3
= 2.7 g / 84.01 g/mol ≈ 0.0321 mol
Since the ratio between NaHCO3 and HCl is 2:1, the number of moles of HCl will be half the number of moles of NaHCO3.
moles of HCl = 0.0321 mol / 2 ≈ 0.0160 mol
Finally, to find the mass of HCl, we multiply the number of moles by the molar mass of HCl. The molar mass of HCl is:
H: 1.01 g/mol
Cl: 35.45 g/mol
Molar mass of HCl = 1.01 + 35.45 = 36.46 g/mol
mass of HCl = moles of HCl * molar mass of HCl
= 0.0160 mol * 36.46 g/mol ≈ 0.583 g
Therefore, 2.7 grams of sodium bicarbonate can neutralize approximately 0.583 grams of hydrochloric acid.