What mass of hydrochloric acid (in grams) can be neutralized by 2.3g of sodium bicarbonate(Hint: Begin by writing a balanced equation for the reaction between aqueous sodium bicarbonate and aqueous hydrochloric acid.)
NaHCO3 + HCl -> NaCl + CO2 + H2O
n(NaHCO3)=n(HCl)
n(NaHCO3)=m/M=2.3g/83.99*(g/mol)=0.027384 mol
m(HCl)=n*M
m(HCl)=0.027384 mol * 36.46 (g/mol)= 0.9984 g
Why did the bicycle fall over? Because it was two-tired! Now, to your question. To find the mass of hydrochloric acid neutralized by sodium bicarbonate, we need to write a balanced equation first. The reaction between sodium bicarbonate (NaHCO3) and hydrochloric acid (HCl) can be represented as:
2 NaHCO3 + 2 HCl -> 2 NaCl + 2 H2O + 2 CO2
According to the equation, for every 2 moles of NaHCO3, we need 2 moles of HCl. Now, let's calculate the molar mass of NaHCO3:
23g (Na) + 1g (H) + 12g (C) + 48g (O) = 84g/mol
Since 2.3g of NaHCO3 is given, let's calculate the number of moles:
2.3g / 84g/mol = 0.0273 mol
Since the reaction is 2:2, we need an equal number of moles of HCl. Therefore:
0.0273 mol HCl
Now, to calculate the mass of HCl, we need to multiply the number of moles by its molar mass. The molar mass of HCl is:
1g (H) + 35.5g (Cl) = 36.5g/mol
So, the mass of HCl needed to neutralize 2.3g of NaHCO3 would be:
0.0273 mol x 36.5g/mol = 0.996g
Therefore, approximately 1g of hydrochloric acid can be neutralized by 2.3g of sodium bicarbonate. Keep the equations balanced, and you'll keep your chemistry more stable!
To find the mass of hydrochloric acid that can be neutralized by 2.3g of sodium bicarbonate, we first need to write a balanced equation for the reaction between sodium bicarbonate (NaHCO3) and hydrochloric acid (HCl):
NaHCO3 + HCl → NaCl + H2O + CO2
From the balanced equation, we can see that 1 mole of sodium bicarbonate reacts with 1 mole of hydrochloric acid.
Next, we need to find the molar mass of sodium bicarbonate. The molar mass is calculated by adding up the atomic masses of all the atoms in a molecule.
NaHCO3 = (1 x atomic mass of Na) + (1 x atomic mass of H) + (1 x atomic mass of C) + (3 x atomic mass of O)
= 23 + 1 + 12 + 48
= 84 g/mol
Now, we can calculate the number of moles of sodium bicarbonate:
Number of moles = mass / molar mass
Number of moles of NaHCO3 = 2.3 g / 84 g/mol
= 0.027 moles
Since the balanced equation shows that 1 mole of sodium bicarbonate reacts with 1 mole of hydrochloric acid, we can conclude that 0.027 moles of hydrochloric acid will be neutralized by 0.027 moles of sodium bicarbonate.
Finally, we need to calculate the mass of hydrochloric acid:
Mass of hydrochloric acid = number of moles x molar mass
Mass of hydrochloric acid = 0.027 moles x (1 x atomic mass of H + 1 x atomic mass of Cl)
Mass of hydrochloric acid = 0.027 moles x (1 + 35.5) g/mol
Mass of hydrochloric acid ≈ 0.972 g
Therefore, approximately 0.972 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 2.3g of sodium bicarbonate, we need to first write a balanced equation for the reaction between sodium bicarbonate (NaHCO3) and hydrochloric acid (HCl).
The balanced equation is:
NaHCO3 + HCl -> NaCl + H2O + CO2
From the equation, we can see that the molar ratio between sodium bicarbonate and hydrochloric acid is 1:1. This means that one mole of sodium bicarbonate reacts with one mole of hydrochloric acid.
To calculate the mass of hydrochloric acid, we'll use the following steps:
Step 1: Calculate the molar mass of sodium bicarbonate (NaHCO3)
The molar mass of Na = 22.99 g/mol
The molar mass of H = 1.01 g/mol
The molar mass of C = 12.01 g/mol
The molar mass of O = 16.00 g/mol
Molar mass of NaHCO3 = (22.99 + 1.01 + (12.01 + 3 * 16.00)) g/mol
= 84.01 g/mol
Step 2: Convert the given mass of sodium bicarbonate to moles
Number of moles of sodium bicarbonate = Given mass / Molar mass
= 2.3 g / 84.01 g/mol
Step 3: Since the molar ratio between sodium bicarbonate and hydrochloric acid is 1:1, the number of moles of hydrochloric acid is also 2.3 g / 84.01 g/mol.
Step 4: Convert the number of moles of hydrochloric acid to mass
Mass of hydrochloric acid = Number of moles x Molar mass
= (2.3 g / 84.01 g/mol) x (1 mol HCl / 1 mol NaHCO3) x (36.46 g/mol HCl)
Using these calculations, the mass of hydrochloric acid that can be neutralized by 2.3g of sodium bicarbonate is approximately 0.993 grams.