you are a manufacturer of antacids. you could use either mg (oh)2 or al(oh)3 in your formulation. you can buy either of these for $2.00 per pound . which of these will provide a better product (i.e., give more relief) for the same amount of money ? explain your answer

To determine which compound, Mg(OH)2 or Al(OH)3, will provide a better product in terms of giving more relief for the same amount of money, we need to consider a few factors.

First, let's compare the molar masses of the compounds to calculate the number of moles we can purchase for $2.00 per pound:

1. Mg(OH)2:
Molar Mass of Mg(OH)2 = 24.31 g/mol (Molar mass of Mg) + (16.00 g/mol (Molar mass of O) + 1.01 g/mol (Molar mass of H)) x 2
Molar Mass of Mg(OH)2 = 58.33 g/mol

2. Al(OH)3:
Molar Mass of Al(OH)3 = 26.98 g/mol (Molar mass of Al) + (16.00 g/mol (Molar mass of O) + 1.01 g/mol (Molar mass of H)) x 3
Molar Mass of Al(OH)3 = 78.00 g/mol

Next, let's calculate the number of moles we can purchase for $2.00 per pound using the molar masses:

1. Mg(OH)2:
Cost of 1 mole of Mg(OH)2 = $2.00/58.33 g/mol = $0.034 per gram

2. Al(OH)3:
Cost of 1 mole of Al(OH)3 = $2.00/78.00 g/mol = $0.026 per gram

Now, we compare the number of moles we can purchase for the two compounds:

1. Mg(OH)2:
Number of moles for $2.00 = $2.00/$0.034 = 58.82 moles

2. Al(OH)3:
Number of moles for $2.00 = $2.00/$0.026 = 76.92 moles

Based on these calculations, we can see that for the same amount of money, we can purchase more moles (approximately 31% more) of Al(OH)3 compared to Mg(OH)2.

Therefore, using Al(OH)3 in the antacid formulation will provide more relief for the same amount of money compared to using Mg(OH)2.