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Whát masŠ of precipitate (In g) Is formed when 20.5 mL of 0.300 Ni(NO3)2 reacts with 33.5 mL of 0.300 M NaOH in the following chemical reaction? Ni(NO3)2(aq) + 2 N2OH(aq) → Ni(OH).(s) + 2 NaNO.(aq)

Whát masŠ of precipitate (In g) Is formed when 20.5 mL of 0.300 Ni(NO3)2 reacts with 33.5 mL of 0.300 M NaOH in the following chemical reaction? Ni(NO3)2(aq) + 2 N2OH(aq) → Ni(OH).(s) + 2 NaNO.(aq)

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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**Chemical Reaction Precipitation Calculation** In this example, we are tasked with determining the mass of the precipitate formed in a chemical reaction. The reaction involves: - 20.5 mL of 0.300 M Nickel(II) nitrate [Ni(NO₃)₂] - 33.5 mL of 0.300 M Sodium hydroxide (NaOH) The balanced chemical equation for the reaction is: \[ \text{Ni(NO}_3\text{)}_2(aq) + 2 \text{NaOH}(aq) \rightarrow \text{Ni(OH)}_2(s) + 2 \text{NaNO}_3(aq) \] Based on this equation, Nickel(II) hydroxide [Ni(OH)₂] is formed as a solid precipitate. **Steps:** 1. **Calculate the Moles of Reactants:** - Use the molarity and volume of each solution to find moles. - \( \text{Moles of Ni(NO}_3\text{)}_2 = 0.300 \, \text{M} \times 0.0205 \, \text{L} \) - \( \text{Moles of NaOH} = 0.300 \, \text{M} \times 0.0335 \, \text{L} \) 2. **Determine the Limiting Reactant:** - Compare stoichiometric ratios to determine which reactant limits the formation of Ni(OH)₂. 3. **Calculate Mass of Precipitate:** - Use the moles of the limiting reactant to find the moles of Ni(OH)₂. - Determine the mass using the molar mass of Ni(OH)₂. An input field is shown at the bottom with a numeric keypad to enter the mass of the precipitate in grams.
Transcribed Image Text:**Chemical Reaction Precipitation Calculation** In this example, we are tasked with determining the mass of the precipitate formed in a chemical reaction. The reaction involves: - 20.5 mL of 0.300 M Nickel(II) nitrate [Ni(NO₃)₂] - 33.5 mL of 0.300 M Sodium hydroxide (NaOH) The balanced chemical equation for the reaction is: \[ \text{Ni(NO}_3\text{)}_2(aq) + 2 \text{NaOH}(aq) \rightarrow \text{Ni(OH)}_2(s) + 2 \text{NaNO}_3(aq) \] Based on this equation, Nickel(II) hydroxide [Ni(OH)₂] is formed as a solid precipitate. **Steps:** 1. **Calculate the Moles of Reactants:** - Use the molarity and volume of each solution to find moles. - \( \text{Moles of Ni(NO}_3\text{)}_2 = 0.300 \, \text{M} \times 0.0205 \, \text{L} \) - \( \text{Moles of NaOH} = 0.300 \, \text{M} \times 0.0335 \, \text{L} \) 2. **Determine the Limiting Reactant:** - Compare stoichiometric ratios to determine which reactant limits the formation of Ni(OH)₂. 3. **Calculate Mass of Precipitate:** - Use the moles of the limiting reactant to find the moles of Ni(OH)₂. - Determine the mass using the molar mass of Ni(OH)₂. An input field is shown at the bottom with a numeric keypad to enter the mass of the precipitate in grams.
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