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Q. An ingestion tablet contains a mass of 0.30 g of hydroxide [Mg(OH)2] as its only basic ingre magnesium dient. The balanced chemical equation between magnesium and hydrochloric acid (HCl(aq)), the acid produced in the stomach is as follows:
Mg(OH)2(aq) + 2HCl(aq) ⟶ MgCl2(aq) + 2H2O(l)
(i) Calculate the volume of 1.00 M HCl neutralised by two of these ingestion tablets?
(ii) What mass of the salt is formed in this neutralisation reaction?
(iii) How many magnesium ions are present in this amount of a salt?

Respuesta :

(i) The volume of  1.00 M HCl neutralized by two of these ingestion tablets is 20.6 mL

(ii) The mass of salt formed in the neutralization reaction is 0.98 g

(iii) The number of magnesium ions present in that amount of salt is 6.19 × 10²¹ ions

Stoichiometry

(i) From the question, we are to calculate the volume of 1.00 M HCl neutralized two of the ingestion tablets

From the given balanced chemical equation

Mg(OH)₂(aq) + 2HCl(aq) ⟶ MgCl₂(aq) + 2H₂O(l)

This means,

1 mole of Mg(OH)₂ is neutralized by 2 moles of HCl to give 1 mole of MgCl₂

Now, we will determine the number of moles of Mg(OH)₂ present in two of the ingestion tablets

From the given information

Mass of Mg(OH)₂ in 1 ingestion tablet = 0.30 g

∴ Mass of Mg(OH)₂ in 2 ingestion tablets = 0.60 g

Using the formula,

[tex]Number\ of\ moles = \frac{Mass}{Molar\ mass}[/tex]

Molar mass of Mg(OH)₂ = 58.33 g/mol

Then,

Number of moles of Mg(OH)₂ in 2 ingestion tablets = [tex]\frac{0.60}{58.33}[/tex]

Number of moles of Mg(OH)₂ in 2 ingestion tablets = 0.0102863 mole

Now,

If 1 mole of Mg(OH)₂ is neutralized by 2 moles of HCl

Then,

0.0102863 mole of Mg(OH)₂ will be neutralized by 0.0205726 moles of HCl

∴ The number of moles of HCl that would neutralize 2 ingestion tablets is 0.0205726 mole

Now, for the volume of 1.00 M HCl neutralized by two ingestion tablets

Using the formula,

[tex]Volume = \frac{Number\ of\ moles}{Concentration}[/tex]

Volume = [tex]\frac{0.0205726}{1.00}[/tex]

Volume = 0.0205726 L

Volume = 20.5726 mL

Volume ≅ 20.6 mL

Hence, the volume of  1.00 M HCl neutralized by two of these ingestion tablets is 20.6 mL

(ii) For the mass of salt (MgCl₂) formed in the neutralization reaction

Since 1 mole of Mg(OH)₂ is neutralized by 2 moles of HCl to give 1 mole of MgCl₂

Then,

0.0102863 mole of Mg(OH)₂ will be neutralized by 0.0205726 moles of HCl  to give 0.0102863  mole of MgCl₂

∴ The number of moles of the salt (MgCl₂) formed is 0.0102863 mole

For the mass of the salt formed,

Using the formula,

Mass = Number of moles × Molar mass

Molar mass of MgCl₂ = 95.211 g/mol

∴ Mass = 0.0102863 × 95.211

Mass = 0.9793689 g

Mass ≅ 0.98 g

Hence, the mass of salt formed in the neutralization reaction is 0.98 g

(iii) For the number of magnesium ions present in that amount of salt,

1 mole of MgCl₂ contains 1 mole of magnesium ions

Then,

0.0102863 mole of MgCl₂ will contain 0.0102863 mole of magnesium ions

From the formula,

Number of ions = Number of moles × Avogadro's constant

Avogadro's constant = 6.022 × 10²³ mol⁻¹

∴ Number of magnesium ions present = 0.0102863 × 6.022 × 10²³

Number of magnesium ions present = 0.061944 × 10²³

Number of magnesium ions present = 6.1944 × 10²¹

Number of magnesium ions present ≅ 6.19 × 10²¹ ions

Hence, the number of magnesium ions present in that amount of salt is 6.19 × 10²¹ ions

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