The approximate concentration of hydrochloric acid HCL, in the stomach (stomach acid) is 0.17M, calculate the mass of the following antacid required to neutralize 50cm of this acid; bicarbonate of soda, NaHCO3
I'm sure you meant 50 cc or 50 cm^3 but NOT 50 cm.
NaHCO3 + HCl ==> NaCl + H2O + CO2
millimoles HCl = mL x M = 50 mL x 0.17 = approx 85
millimoles NaHCO3 neutralized = approx 85 = 0.085 moles
grams NaHCO3 = moles NaHCO3 x molar mass NaHCO3 = ? grams.
To calculate the mass of bicarbonate of soda (NaHCO3) required to neutralize 50 cm³ of hydrochloric acid (HCl), we need to use the concept of stoichiometry.
Before we proceed, we need to know the balanced chemical equation for the reaction between HCl and NaHCO3:
HCl + NaHCO3 → NaCl + H2O + CO2
From the balanced equation, we can see that the ratio of HCl to NaHCO3 is 1:1. This means that 1 mole of HCl reacts with 1 mole of NaHCO3.
Step 1: Calculate the number of moles of HCl in 50 cm³ of 0.17 M HCl.
To do this, we use the formula:
moles = volume (in liters) × concentration (in moles per liter)
We convert 50 cm³ to liters by dividing by 1000:
volume = 50 cm³ / 1000 = 0.05 L
Now we can calculate the moles of HCl:
moles of HCl = 0.05 L × 0.17 mol/L = 0.0085 moles
Step 2: Calculate the mass of NaHCO3 required using the mole ratio.
Since the mole ratio between HCl and NaHCO3 is 1:1, we know that 0.0085 moles of NaHCO3 are required.
To calculate the mass, we use the formula:
mass = moles × molar mass
The molar mass of NaHCO3 can be found by adding up the atomic masses of each element:
Na: 22.99 g/mol
H: 1.01 g/mol
C: 12.01 g/mol
O: 16.00 g/mol
Molar mass of NaHCO3 = (22.99 g/mol) + (1.01 g/mol) + (12.01 g/mol) + (3 × 16.00 g/mol) = 84.01 g/mol
Now we can calculate the mass of NaHCO3:
mass = 0.0085 moles × 84.01 g/mol ≈ 0.714 g
Therefore, approximately 0.714 grams of bicarbonate of soda (NaHCO3) is required to neutralize 50 cm³ of 0.17 M hydrochloric acid (HCl).
To calculate the mass of bicarbonate of soda (NaHCO3) required to neutralize 50 cm³ of hydrochloric acid (HCl), we can use the balanced chemical equation for the reaction:
HCl + NaHCO3 → NaCl + H2O + CO2
From the equation, we can see that the stoichiometry between HCl and NaHCO3 is 1:1. This means that one mole of HCl reacts with one mole of NaHCO3.
First, let's calculate the number of moles of HCl using the given concentration and volume:
moles of HCl = concentration × volume
moles of HCl = 0.17 M × 50 cm³
Next, since the mole ratio is 1:1, the number of moles of NaHCO3 required to neutralize HCl will also be the same as the moles of HCl.
Now, we need to calculate the molar mass of NaHCO3:
molar mass of NaHCO3 = (atomic mass of Na) + (atomic mass of H) + (atomic mass of C) + 3 × (atomic mass of O)
molar mass of NaHCO3 = (22.99 g/mol) + (1.01 g/mol) + (12.01 g/mol) + 3 × (16.00 g/mol)
molar mass of NaHCO3 = 22.99 g/mol + 1.01 g/mol + 12.01 g/mol + 48.00 g/mol
molar mass of NaHCO3 = 84.02 g/mol
Finally, we can calculate the mass of NaHCO3 using the moles of NaHCO3 and its molar mass:
mass of NaHCO3 = moles of NaHCO3 × molar mass of NaHCO3
Therefore, the mass of bicarbonate of soda (NaHCO3) required to neutralize 50 cm³ of hydrochloric acid (HCl) is given by:
mass of NaHCO3 = (0.17 M × 50 cm³) × (84.02 g/mol)