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What is the volume of HCl (r= 1.17 g / cm ^ 3, omega = 36 \%) required to precipitate silver in the form of AgCl from 2, 000 g of an alloy containing 22% by weight of Ag, if used 1.5 times the amount of precipitating reagent?​

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mickymike92

The volume of precipitating reagent, HCl required to precipitate silver in the form of AgCl from 2,000 g of an alloy containing 22% by weight of Ag is 0.530 L

In order to determine the volume of precipitating reagent, HCl required to precipitate silver in the form of AgCl from 2,000 g of an alloy containing 22% by weight of Ag, the following steps are followed:

Step 1.

The concentration of the HCl solution is calculated using the formula:

  • Molarity = percentage * specific gravity * 1000 cm³ / molar mass * 1 L

Percentage purity of HCl = 36% = 0.36

specific gravity, r = 1.17 g/cm³

molar mass of HCl = 36.5 g/mol

Concentration of HCl = 0.36 * 1.17 * 1000 / 36.5 * 1

Concentration of HCl = 11.53 mol/L

Therefore, 1 L of HCl solution contains 11.53 moles of HCl

Step 2:

The equation of the dissociation of HCl solution is given below:

  • HCl (aq) ------> H⁺ (aq) + Cl⁻ (aq)

1 mole of HCl produces 1 mole of Chloride ions, Cl⁻

Equation of reaction between Silver in the alloy and HCl is given as follows:

  • Ag (s) + H⁺Cl⁺ (aq) -----> AgCl (s)

A white precipitate of AgCl is formed but hydrogen gas is not formed as Silver is a less reactive metal.

1 mole of HCl produces 1 mole of Cl⁻ which reacts with 1 mole of Ag to produce 1 mole of AgCl precipitate.

Step 3: Determine the mass and number of moles of silver in the alloy

  • Percentage of silver in the alloy = 22% = 0.22

Mass of silver in 2000 g alloy = 0.22 * 2000 g

Mass of silver in 2000 g alloy = 440 g

Number of moles of silver in 440 g of silver is determined using the formula below:

  • number of moles = mass / molar mass

molar mass of silver = 108 g/mol

number of moles of silver = 440 /108

number of moles of silver in the alloy = 4.074 moles

Step 4: Calculate the volume of HCl required

Since 1 mole of Cl⁻ ions react with 1 mole of Ag; 4.074 moles of Cl⁻ ions will be required to react with 4.074 moles of Ag.

1 L of HCl solution contains 11.53 moles of Cl⁻

Volume of HCl solution that contains 4.074 moles of Cl⁻ = 4.074 * 1L/11.53

Volume of HCl that contains 4.074 moles of Cl⁻ = 0.353 L

Since 1.5 times the amount of precipitating reagent HCl is required, volume of HCl required = 1.5 * 0.353 L

Volume of precipitating reagent HCl required = 0.530 L

Therefore, the volume of precipitating reagent, HCl required to precipitate silver in the form of AgCl from 2,000 g of an alloy containing 22% by weight of Ag is 0.530 L

Learn more about precipitation reactions at: https://brainly.com/question/24944690

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