The approximate concentration of hydrochloric acid in the stomach(stomach acid) is 0.17m.calculate the mass of the following to neutralise 50cm of stomach acid.a)bicarbonate of soda b)aluminium hydro xide
To calculate the mass of bicarbonate of soda (sodium bicarbonate) or aluminium hydroxide required to neutralize 50 cm³ of stomach acid with a concentration of 0.17 M, we need to use the balanced chemical equations for the neutralization reactions.
The balanced chemical equation for the reaction between hydrochloric acid (HCl) and bicarbonate of soda (NaHCO₃) is:
HCl + NaHCO₃ -> NaCl + CO₂ + H₂O
From the balanced equation, we can see that one mole of HCl reacts with one mole of NaHCO₃. The molar mass of NaHCO₃ is approximately 84 g/mol.
To calculate the mass of NaHCO₃ needed, we need to convert the volume of stomach acid into moles using the molar concentration (Molarity) of hydrochloric acid:
Molarity = moles/volume (in liters)
Given that 50 cm³ = 50/1000 = 0.05 L and the concentration of HCl is 0.17 M, we can calculate the moles of HCl:
moles of HCl = Molarity x volume
= 0.17 M x 0.05 L
= 0.0085 moles
Since the mole ratio between HCl and NaHCO₃ is 1:1, the moles of NaHCO₃ needed to neutralize the given amount of HCl is also 0.0085 moles.
Now, let's calculate the mass of NaHCO₃:
mass = moles x molar mass
= 0.0085 moles x 84 g/mol
≈ 0.714 g
Therefore, approximately 0.714 grams of bicarbonate of soda (sodium bicarbonate) are required to neutralize 50 cm³ of stomach acid with a concentration of 0.17 M.
For aluminium hydroxide (Al(OH)₃), the balanced chemical equation for the reaction with HCl is:
3HCl + Al(OH)₃ -> AlCl₃ + 3H₂O
Using the same process as above, you can calculate the moles and mass of aluminium hydroxide needed to neutralize the acid.
To calculate the mass of bicarbonate of soda and aluminum hydroxide required to neutralize 50 cm³ of stomach acid, we need to first determine the balanced chemical equations for the neutralization reactions.
1. Neutralization reaction between hydrochloric acid (HCl) and sodium bicarbonate (NaHCO₃):
HCl + NaHCO₃ → NaCl + CO₂ + H₂O
2. Neutralization reaction between hydrochloric acid (HCl) and aluminum hydroxide (Al(OH)₃):
HCl + Al(OH)₃ → AlCl₃ + H₂O
Now, let's calculate the mass of each substance individually:
a) Bicarbonate of soda (NaHCO₃):
Since the balanced equation tells us that the molar ratio of HCl to NaHCO₃ is 1:1, it means that 1 mole of HCl reacts with 1 mole of NaHCO₃.
Given concentration of HCl = 0.17 M, which means 0.17 moles of HCl in 1 liter (1000 cm³) of stomach acid.
To neutralize 50 cm³ of stomach acid, we need to convert the volume to moles:
50 cm³ x (1 L / 1000 cm³) = 0.05 L
Number of moles of HCl = concentration x volume:
Moles of HCl = 0.17 M x 0.05 L = 0.0085 moles HCl
Since the molar ratio of HCl to NaHCO₃ is 1:1, we need 0.0085 moles of NaHCO₃ to neutralize the acid.
Now, we can calculate the mass of NaHCO₃ using its molar mass (approx. 84 g/mol):
Mass of NaHCO₃ = moles of NaHCO₃ x molar mass of NaHCO₃
Mass of NaHCO₃ = 0.0085 moles x 84 g/mol ≈ 0.71 grams
Therefore, approximately 0.71 grams of bicarbonate of soda (NaHCO₃) is needed to neutralize 50 cm³ of stomach acid.
b) Aluminum hydroxide (Al(OH)₃):
Since the balanced equation tells us that the molar ratio of HCl to Al(OH)₃ is 3:1, it means that 3 moles of HCl react with 1 mole of Al(OH)₃.
Using the same calculations as before, we found that the number of moles of HCl is 0.0085 moles.
To neutralize this amount of HCl, we need 1/3 of that amount of Al(OH)₃:
Moles of Al(OH)₃ = 0.0085 moles / 3 = 0.0028 moles Al(OH)₃
Now, we can calculate the mass of Al(OH)₃ using its molar mass (approx. 78 g/mol):
Mass of Al(OH)₃ = moles of Al(OH)₃ x molar mass of Al(OH)₃
Mass of Al(OH)₃ = 0.0028 moles x 78 g/mol ≈ 0.22 grams
Therefore, approximately 0.22 grams of aluminum hydroxide (Al(OH)₃) is needed to neutralize 50 cm³ of stomach acid.
Please note that these calculations are approximate and actual factors may vary slightly.