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HCl is produced in the stomach which can be neutralised by Mg(OH)2 in the form of milk of magnesia. How much Mg(OH)2 is required to neutralise one mole of stomach acid?
  • a)
    29.16 g
  • b)
    34.3 g
  • c)
    58.33 g
  • d)
    68.66 g
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
HCl is produced in the stomach which can be neutralised by Mg(OH)2 in ...
Mg(OH)2 + 2HCl → MgCl2 + 2H2O
No. of moles of Mg(OH)2 required for 2 moles of HCl = 1
No. of moles of Mg(OH)2 required for 1 mole of HCl = 0.5
Mass of 0.5 mol of Mg(OH)2 = 58.33 x 0.5 = 29.16 g
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Most Upvoted Answer
HCl is produced in the stomach which can be neutralised by Mg(OH)2 in ...
Neutralization Reaction:
The neutralization reaction between hydrochloric acid (HCl) and magnesium hydroxide (Mg(OH)2) can be represented as follows:
HCl + Mg(OH)2 → MgCl2 + 2H2O

Molar Ratios:
From the balanced chemical equation above, we can see that one mole of HCl reacts with one mole of Mg(OH)2.

Molar Mass:
The molar mass of HCl is 36.46 g/mol (1 hydrogen atom + 1 chlorine atom).
The molar mass of Mg(OH)2 is 58.33 g/mol (1 magnesium atom + 2 oxygen atoms + 2 hydrogen atoms).

Calculating Amount of Mg(OH)2:
To calculate the amount of Mg(OH)2 required to neutralize one mole of HCl, we need to use the molar ratio between HCl and Mg(OH)2.

The molar mass of HCl is 36.46 g/mol.
Therefore, to neutralize one mole of HCl, we need 36.46 g of Mg(OH)2.

Final Answer:
The correct answer is option A) 29.16 g.
This can be calculated by dividing the molar mass of HCl (36.46 g/mol) by the molar mass of Mg(OH)2 (58.33 g/mol) and multiplying by the molar mass of Mg(OH)2 (58.33 g/mol):
36.46 g/mol / 58.33 g/mol × 58.33 g/mol = 36.46 g.

Therefore, 36.46 g of Mg(OH)2 is required to neutralize one mole of HCl.
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Community Answer
HCl is produced in the stomach which can be neutralised by Mg(OH)2 in ...
Mg(OH)2 + 2HCl → MgCl2 + 2H2O
No. of moles of Mg(OH)2 required for 2 moles of HCl = 1
No. of moles of Mg(OH)2 required for 1 mole of HCl = 0.5
Mass of 0.5 mol of Mg(OH)2 = 58.33 x 0.5 = 29.16 g
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HCl is produced in the stomach which can be neutralised by Mg(OH)2 in the form of milk of magnesia. How much Mg(OH)2 is required to neutralise one mole of stomach acid?a)29.16 gb)34.3 gc)58.33 gd)68.66 gCorrect answer is option 'A'. Can you explain this answer?
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