The approximate concentration of hydrochloric acid. HCL in stomach is 0.17M. Calculate the mass of the following anatcids bicarbonate of soda NaHCO3
grams of anacid= .17(volumestomach)*molmassAnacid
To calculate the mass of NaHCO3 needed to neutralize the hydrochloric acid (HCl) in the stomach, we can use the stoichiometry of the balanced chemical equation between HCl and NaHCO3.
The balanced equation is as follows:
HCl + NaHCO3 -> NaCl + CO2 + H2O
From the balanced equation, we can see that for every 1 mole of HCl, we need 1 mole of NaHCO3 to completely neutralize the acid.
Given that the concentration of HCl in the stomach is 0.17 M (0.17 moles per liter), we need to convert this to moles of HCl.
Volume of HCl (V) = 1 L (assuming 1 liter of stomach acid)
Concentration of HCl (C) = 0.17 M
Number of moles of HCl (n) = C x V
n = 0.17 M x 1 L = 0.17 moles
Since the stoichiometry of the reaction tells us that we need an equal number of moles of NaHCO3 to neutralize the HCl, we also need 0.17 moles of NaHCO3.
The molar mass of NaHCO3 is calculated as follows:
Na = 22.99 g/mol
H = 1.008 g/mol
C = 12.01 g/mol
O = 16.00 g/mol (3 oxygens in NaHCO3)
Molar mass of NaHCO3 = (22.99 + 1.008 + 12.01 + (3 x 16.00)) g/mol
Molar mass of NaHCO3 ≈ 84.01 g/mol
Now we can calculate the mass of NaHCO3 needed:
Mass of NaHCO3 = Number of moles x Molar mass
Mass of NaHCO3 = 0.17 moles x 84.01 g/mol
Therefore, the mass of NaHCO3 needed to neutralize the hydrochloric acid in the stomach is approximately 14.28 grams.
To calculate the mass of NaHCO3 (bicarbonate of soda) required to neutralize 0.17M hydrochloric acid (HCl), you need to use the balanced chemical equation for the reaction between HCl and NaHCO3.
The balanced chemical equation is as follows:
HCl + NaHCO3 → NaCl + CO2 + H2O
From the equation, we can see that 1 mole of HCl reacts with 1 mole of NaHCO3 to produce 1 mole of NaCl, 1 mole of CO2, and 1 mole of H2O.
To convert from moles to mass, you need to know the molar mass of NaHCO3, which can be calculated by adding together the atomic masses of each element in the compound:
Molar mass of NaHCO3 = (atomic mass of Na) + (atomic mass of H) + (atomic mass of C) + 3 x (atomic mass of O)
Atomic masses:
Na: 22.99 g/mol
H: 1.01 g/mol
C: 12.01 g/mol
O: 16.00 g/mol
Therefore, Molar mass of NaHCO3 = 22.99 + 1.01 + 12.01 + 3 x 16.00 = 84.01 g/mol
Now, let's calculate the mass of NaHCO3 needed to neutralize the HCl:
1. Calculate the number of moles of HCl using the given concentration and the volume of the solution:
moles of HCl = concentration x volume
Given concentration: 0.17M
Volume is not provided, so let's assume it to be 1 liter for simplicity.
moles of HCl = 0.17 mol/L x 1 L = 0.17 mol
2. Since the balanced equation shows a 1:1 stoichiometric ratio between HCl and NaHCO3, the number of moles of NaHCO3 needed is also 0.17 mol.
3. Finally, calculate the mass of NaHCO3 using the molar mass:
mass of NaHCO3 = moles of NaHCO3 x molar mass of NaHCO3
mass of NaHCO3 = 0.17 mol x 84.01 g/mol = 14.28 grams
Therefore, approximately 14.28 grams of NaHCO3 (bicarbonate of soda) is needed to neutralize 0.17M hydrochloric acid.