Solution Stoichiometry: Practice Questions Instructions Answer the following questions in the space provided. 1. What mass of gold is produced from the reaction of excess iron with 150 mL of 0.75 mol/L gold(III) nitrate solution?

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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Solution Stoichiometry: Practice Questions
Instructions
Answer the following questions in the space provided.
What mass of gold is produced from the reaction of excess iron with 150 mL of 0.75 mol/L gold (1II) nitrate
solution?
Transcribed Image Text:Solution Stoichiometry: Practice Questions Instructions Answer the following questions in the space provided. What mass of gold is produced from the reaction of excess iron with 150 mL of 0.75 mol/L gold (1II) nitrate solution?
Expert Solution
Step 1: Writing the given information,

First we have to write the balanced chemical equation between the iron and gold(III) nitrate

Fe + Au(NO3)3 → Fe(NO3)2 + 2Au

Balancing the above equation, we get

3Fe + 2Au(NO3)3 → 3Fe(NO3)2 + 2Au

Excess reactant: Fe

Concentration of gold (III) nitrate solution = 0.75 mol/L

Volume of gold (III) nitrate solution = 150 mL = 0.150 L

Note:

Molar mass of gold = 196.97 g/mol

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