Chemistry, 15.04.2020 01:46 jaileen84

A student has a mixture of salt (NaCl) and sugar (C12H22O11). To determine the percent composition, the student measures out 5.84 g of the mixture and dissolves the sample in 100.0 mL of water. The student knows aqueous NaCl will react with aqueous silver nitrate (AgNO3) to form solid silver chloride (AgCl). The student determines the sugar will not react with silver nitrate. Write a balanced net ionic equation for the reaction of NaCl with AgNO3. How many mL of 1.0 M AgNO3 will be required to precipitate 5.84 g of AgCl

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Answer from: campbelldean

1. Net ionic equation:

Cl⁻(aq) + Ag⁺(aq) → AgCl(s)

2. Volume of 1.0M AgNO₃

41ml

Explanation:

1. Net ionic equation for the reaction of NaCl with AgNO₃.

i) Molecular equation:

It is important to show the phases:

(aq) for ions in aqueous solution(s) for solid compounds or elements(g) for gaseous compounds or elementsNaCl(aq) + AgNO₃(aq) → AgCl(s) + NaNO₃(aq)

ii) Dissociation reactions:

Determine the ions formed:

NaCl(aq) → Na⁺(aq) + Cl⁻(aq)AgNO₃(aq) → Ag⁺(aq) + NO₃⁻(aq)NaNO₃(aq) → Na⁺(aq) + NO₃⁻(aq)

iii) Total ionic equation:

Substitute the aqueous compounds with the ions determined above:

Na⁺(aq) + Cl⁻(aq) + Ag⁺(aq) + NO₃⁻(aq) → AgCl(s) + Na⁺(aq) + NO₃⁻(aq)

iv) Net ionic equation

Remove the spectator ions:

Cl⁻(aq) + Ag⁺(aq) → AgCl(s) ← answer

2. How many mL of 1.0 M AgNO₃ will be required to precipitate 5.84 g of AgCl

i) Determine the number of moles of AgNO₃

The reaction is 1 to 1: 1 mole of AgNO₃ produces 1 mol of AgCl

The number of moles of AgCl is determined using the molar mass:

number of moles = mass in grams / molar massmolar mass of AgCl = 143.32g/molnumber of moles = 5.84g / (143.32g/mol) = 0.040748 mol

ii) Determine the volume of AgNO₃

molarity = number of moles of solute / volume of solution in liters1.0M = 0.040748mol / VV = 0.040748mol / (1.0M) = 0.040748 literV = 0.040748liter × 1,000ml / liter = 40.748 ml

Round to two significant figures: 41ml ← answer

Answer from: Quest

Ifound this on google “these two compounds have a different ionic structure. caesium ion is larger in size when compared to the sodium ion. the charge gets more area to spread out evenly. so caesium chloride has a lower melting point than sodium chloride.” i hope this you for what your looking for

Answer from: Quest

it forms an anion that has a larger radius. anions have a negative charge and are larger than the original atom.

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A student has a mixture of salt (NaCl) and sugar (C12H22O11). To determine the percent composition,...

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