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A precipitate is formed after an iron(III) nitrate solution is mixed with a sodium hydroxide solution. The volume of each solution is 25.0 mL and the concentration of each solution is 0.10 M. How many grams of precipitate would you expect to form from this reaction?

Respuesta :

Answer:

0.2675 g.

Explanation:

Equation of the reaction

Fe(NO3)3(aq) + 3NaOH(aq) --> Fe(OH)3(s) + 3NaNO3(aq)

Calculating the number of moles:

Fe(NO3)3 and NaOH:

Number of moles = molar concentration * volume

= 0.025*0.1

= 0.0025 mol.

By stoichiometry, since 1 mole of the reactants each will react to form 1 mole of Fe(OH)3

Therefore, number of moles of Fe(OH)3 = 0.0025 mol

Calculating molar mass of Fe(OH)3

= 56 + (16 + 1)*3

= 107 g/mol.

mass of Fe(OH)3 = molar mass * number of moles

= 107 * 0.0025

= 0.2675 g of precipitate, Fe(OH)3.

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