According to the following reaction, how many moles of iron(II) nitrate will be formed upon the complete reaction of 0.676 moles iron(II) chloride with excess silver nitrate? FeCl (aq) + 2AGNO, (aq) → Fe(NOs)2 (aq) + 2AgCl(s) mol iron(II) nitrate
According to the following reaction, how many moles of iron(II) nitrate will be formed upon the complete reaction of 0.676 moles iron(II) chloride with excess silver nitrate? FeCl (aq) + 2AGNO, (aq) → Fe(NOs)2 (aq) + 2AgCl(s) mol iron(II) nitrate
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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![**Problem Statement:**
**Chemical Reaction Stoichiometry**
**Objective:** According to the following reaction, how many moles of iron(II) nitrate will be formed upon the complete reaction of 0.676 moles of iron(II) chloride with excess silver nitrate?
**Given Reaction:**
\[ \text{FeCl}_2(\text{aq}) + 2\text{AgNO}_3(\text{aq}) \rightarrow \text{Fe(NO}_3\text{)}_2(\text{aq}) + 2\text{AgCl}(\text{s}) \]
**Calculation:**
_Enter the number of moles:_ \( \boxed{} \) _mol iron(II) nitrate_
The balanced chemical equation provides the stoichiometric ratios of the reactants and products. From the equation, 1 mole of iron(II) chloride (FeCl₂) reacts with 2 moles of silver nitrate (AgNO₃) to produce 1 mole of iron(II) nitrate (Fe(NO₃)₂) and 2 moles of silver chloride (AgCl).
Use the stoichiometric relationship:
- 1 mole FeCl₂ → 1 mole Fe(NO₃)₂
Therefore, if 0.676 moles of FeCl₂ are used, the moles of Fe(NO₃)₂ formed will be:
\[ 0.676 \text{ mol FeCl}_2 \times \frac{1 \text{ mol Fe(NO}_3\text{)}_2}{1 \text{ mol FeCl}_2} = 0.676 \text{ mol Fe(NO}_3\text{)}_2 \]
The answer should be entered into the provided box as:
\[ 0.676 \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1f61489d-b599-4717-b8e4-1c0e8dbf11ad%2F0c4b39a9-f4aa-46ae-b627-3816fd521743%2Fmy1aii.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
**Chemical Reaction Stoichiometry**
**Objective:** According to the following reaction, how many moles of iron(II) nitrate will be formed upon the complete reaction of 0.676 moles of iron(II) chloride with excess silver nitrate?
**Given Reaction:**
\[ \text{FeCl}_2(\text{aq}) + 2\text{AgNO}_3(\text{aq}) \rightarrow \text{Fe(NO}_3\text{)}_2(\text{aq}) + 2\text{AgCl}(\text{s}) \]
**Calculation:**
_Enter the number of moles:_ \( \boxed{} \) _mol iron(II) nitrate_
The balanced chemical equation provides the stoichiometric ratios of the reactants and products. From the equation, 1 mole of iron(II) chloride (FeCl₂) reacts with 2 moles of silver nitrate (AgNO₃) to produce 1 mole of iron(II) nitrate (Fe(NO₃)₂) and 2 moles of silver chloride (AgCl).
Use the stoichiometric relationship:
- 1 mole FeCl₂ → 1 mole Fe(NO₃)₂
Therefore, if 0.676 moles of FeCl₂ are used, the moles of Fe(NO₃)₂ formed will be:
\[ 0.676 \text{ mol FeCl}_2 \times \frac{1 \text{ mol Fe(NO}_3\text{)}_2}{1 \text{ mol FeCl}_2} = 0.676 \text{ mol Fe(NO}_3\text{)}_2 \]
The answer should be entered into the provided box as:
\[ 0.676 \]
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