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Suppose one magnesium hydroxide tablet costs 0.0500 dollars. Let each 1.00 g magnesium hydroxide tablet be 100. percent magnesium hydroxide. Using only these tablets, you are required to neutralize 2.00 L of 0.500 M HCl. How much does this cost? Express your answer in dollars.

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superman1987

Answer:

1.46 $.

Explanation:

  • Mg(OH)₂ neutralizes HCl according to the balanced reaction:

2HCl + Mg(OH)₂ → MgCl₂ + 2H₂O.

Every 2 moles of HCl are neutralized by 1 mole of Mg(OH)₂.

  • At neutralization: The no. of moles of HCl is equal to the no. of moles of Mg(OH)₂.

The no. of moles of HCl = (MV) of HCl = (2.0 L)(0.5 mol/L) = 1.0 mol.

This requires 0.5 mol of Mg(OH)₂ to be neutralized.

∵ no. of moles of Mg(OH)₂ = mass/molar mass.

∴ mass of Mg(OH)₂ = (no. of moles of Mg(OH)₂)(molar mass) = (0.5 mol)(58.32 g/mol) = 29.16 g.

  • To find the cost, we can use cross multiplication:

1.0 g of Mg(OH)₂ costs → 0.05 $.

29.16 g of Mg(OH)₂ costs → ??? $.

∴ The cost of tablets = (29.16 g)(0.05 $) = 1.46 $.

Oseni

The cost of the magnesium hydroxide tablet that will be required to neutralize 2.00 L of 0.500 M HCl would be 1.46 dollars

Stoichiometric equation

From the equation of the reaction:

Mg(OH)2 + 2HCl ---------------> MgCl2 + 2H2O

The mole ratio of Mg(OH)2 to HCl = 1:2

Mole of 2.00 L, 0.500 M HCl = 0.500 x 2.00 = 1 mole

Equivalent mole of Mg(OH)2 = 1/2 = 0.5 moles

Mass of 0.5 mole Mg(OH)2 = 0.5 x 58.3

                                                  = 29.15 g

Since 1 tablet = 1.00 g Mg(OH)2 = 0.05 dollars.

29.15 g Mg(OH)2 = 29.15 tablet = 29.15 x 0.05 dollars

                                      = 1.46 dollars

More on stoichiometric calculations can be found here: https://brainly.com/question/8062886

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