What mass of hydrochloric acid (in grams) can be neutralized by 2.3g of sodium bicarbonate? (: Begin by writing a balanced equation for the reaction between aqueous sodium bicarbonate and aqueous hydrochloric acid.)
first write the chemical equation:
HCl + NaHCO3 -> NaCl + H2O + CO2
observe that it's already balanced. :)
now we get the molar mass of NaHCO3 (sodium bicarbonate) and HCl (hydrochloric acid),, i expect you already know how to find molar masses.
NaHCO3 : 23 + 1 + 12 + 3*16 = 84 g/mol
HCl = 1 + 35.45 = 36.45 g/mol
*note: i used approximate mass of each element.
solving,
2.3 g NaHCO3 * ( 1 mol NaHCO3 / 84 g NaHCO3 ) * (1 mol HCl / 1 mol NaHCO3) * (36.45 g HCl / 1 mol HCl) = 0.998 g HCl
hope this helps~ :)
Why did the sodium bicarbonate go to therapy? Because it felt a bit unbalanced!
But don't worry, I've got you covered. The balanced equation for the reaction between aqueous sodium bicarbonate (NaHCO3) and aqueous hydrochloric acid (HCl) is:
NaHCO3 + HCl → NaCl + H2O + CO2
Now that we have a balanced equation, let's calculate the mass of hydrochloric acid that can be neutralized by 2.3g of sodium bicarbonate.
The molar mass of NaHCO3 is 84.01 g/mol, and from the equation, we know that 1 mole of NaHCO3 reacts with 1 mole of HCl.
Therefore, 2.3g of NaHCO3 is equal to:
2.3g / 84.01 g/mol = 0.0274 mol of NaHCO3.
Since the ratio is 1:1, this means that 0.0274 mol of HCl will be neutralized.
Now, let's calculate the mass of HCl assuming the molar mass of HCl is 36.46 g/mol:
0.0274 mol x 36.46 g/mol = 1 g (approximately).
So, approximately 1 g of hydrochloric acid can be neutralized by 2.3g of sodium bicarbonate. Keep in mind that these values are approximate due to rounding.
The balanced equation for the reaction between aqueous sodium bicarbonate (NaHCO3) and aqueous hydrochloric acid (HCl) is:
NaHCO3 + HCl → NaCl + H2O + CO2
To find the mass of hydrochloric acid that can be neutralized by 2.3g of sodium bicarbonate, we need to determine the stoichiometry of the reaction.
From the balanced equation, we can see that for every 1 mole of sodium bicarbonate (NaHCO3), it reacts with 1 mole of hydrochloric acid (HCl).
The molar mass of sodium bicarbonate is:
Na: 22.99 g/mol
H: 1.01 g/mol
C: 12.01 g/mol
O: 16.00 g/mol
So, the molar mass of NaHCO3 is:
22.99 + 1.01 + 12.01 + (3 * 16.00) = 84.01 g/mol
Now, we can calculate the number of moles of sodium bicarbonate:
moles = mass / molar mass = 2.3g / 84.01 g/mol ≈ 0.0274 mol
Since the stoichiometry of the balanced equation is 1:1, the number of moles of hydrochloric acid neutralized will also be 0.0274 mol.
Now, we can find the mass of hydrochloric acid:
mass = moles * molar mass = 0.0274 mol * (1.01 g/mol) = 0.0277 g
Therefore, approximately 0.0277 grams of hydrochloric acid can be neutralized by 2.3 grams of sodium bicarbonate.
To determine the mass of hydrochloric acid that can be neutralized by sodium bicarbonate, we need to write a balanced chemical equation and then use stoichiometry.
The balanced equation for the reaction between sodium bicarbonate (NaHCO3) and hydrochloric acid (HCl) is:
NaHCO3 + HCl -> NaCl + H2O + CO2
From the equation, we see that 1 mole of NaHCO3 reacts with 1 mole of HCl, producing 1 mole of NaCl, 1 mole of H2O, and 1 mole of CO2.
The molar mass of NaHCO3 is calculated as follows:
23.0 g/mol (Na) + 1.0g/mol (H) + 12.0 g/mol (C) + 48.0g/mol (O) = 84.0 g/mol
So, 1 mole of NaHCO3 weighs approximately 84.0 grams.
To find the mass of HCl required, we need to use stoichiometry. Since 2.3 grams of NaHCO3 is provided, we can calculate the number of moles:
Moles of NaHCO3 = Mass of NaHCO3 / Molar mass of NaHCO3
= 2.3 g / 84.0 g/mol
≈ 0.027 moles
Since the balanced equation shows a 1:1 ratio between NaHCO3 and HCl, it means that 0.027 moles of HCl are needed to neutralize the given amount of NaHCO3.
Finally, to find the mass of HCl:
Mass of HCl = Moles of HCl * Molar mass of HCl
The molar mass of HCl is approximately 36.5 g/mol.
Mass of HCl = 0.027 moles * 36.5 g/mol
≈ 0.9855 grams
Therefore, approximately 0.9855 grams of hydrochloric acid can be neutralized by 2.3 grams of sodium bicarbonate.