Question:
How many grams of \(Mg\) is required to completely reduce \(100\;ml,\;0.1\;M\;NO_3^-\) solution using the following reaction?
\[
NO_3^- + Mg \longrightarrow Mg^{2+} + NH_3
\]
How many grams of \(Mg\) is required to completely reduce \(100\;ml,\;0.1\;M\;NO_3^-\) solution using the following reaction?
\[ NO_3^- + Mg \longrightarrow Mg^{2+} + NH_3 \]
\[ NO_3^- + Mg \longrightarrow Mg^{2+} + NH_3 \]
Show Hint
For redox stoichiometry, first calculate the change in oxidation number. Then compare electrons gained and lost to find the mole ratio between oxidizing and reducing agents.
Updated On: Jun 22, 2026
- \(0.96\)
- \(0.62\)
- \(0.24\)
- \(0.75\)
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The Correct Option is A
Solution and Explanation
Step 1: Identify the oxidation and reduction changes.
In the reaction, nitrate ion \(NO_3^-\) is reduced to ammonia \(NH_3\).
The oxidation state of nitrogen in \(NO_3^-\) is calculated as:
\[ x+3(-2)=-1 \] \[ x-6=-1 \] \[ x=+5 \] So, nitrogen has oxidation state \(+5\) in \(NO_3^-\).
In \(NH_3\), hydrogen has oxidation state \(+1\). Therefore, oxidation state of nitrogen is:
\[ x+3(+1)=0 \] \[ x+3=0 \] \[ x=-3 \] So, nitrogen has oxidation state \(-3\) in \(NH_3\).
Step 2: Calculate the number of electrons gained by nitrate ion.
Nitrogen changes from \(+5\) to \(-3\).
Therefore, decrease in oxidation state is:
\[ +5-(-3)=8 \] Hence, one \(NO_3^-\) ion gains \(8\) electrons during reduction.
\[ NO_3^- + 8e^- \longrightarrow NH_3 \]
Step 3: Calculate the electrons lost by magnesium.
Magnesium changes from \(Mg\) to \(Mg^{2+}\).
So, magnesium loses \(2\) electrons:
\[ Mg \longrightarrow Mg^{2+}+2e^- \] Thus, \(1\) mole of \(Mg\) gives \(2\) moles of electrons.
Step 4: Find the mole ratio between \(NO_3^-\) and \(Mg\).
One mole of \(NO_3^-\) requires \(8\) moles of electrons.
One mole of \(Mg\) supplies \(2\) moles of electrons.
Therefore, moles of \(Mg\) required for \(1\) mole of \(NO_3^-\) are:
\[ \frac{8}{2}=4 \] So, the balanced electron relation is:
\[ 1\;mole\;NO_3^- \equiv 4\;moles\;Mg \]
Step 5: Calculate moles of nitrate ion present.
Given volume of \(NO_3^-\) solution is:
\[ 100\;ml = 0.100\;L \] Molarity of \(NO_3^-\) solution is:
\[ 0.1\;M \] Moles of \(NO_3^-\) are calculated by:
\[ \text{Moles} = Molarity \times Volume\;in\;litres \] \[ \text{Moles of }NO_3^- = 0.1 \times 0.100 \] \[ =0.010\;mol \]
Step 6: Calculate moles of magnesium required.
Since \(1\) mole of \(NO_3^-\) requires \(4\) moles of \(Mg\),
\(0.010\) mole of \(NO_3^-\) will require:
\[ 0.010 \times 4 = 0.040\;mol \] Thus, required moles of magnesium are:
\[ 0.040\;mol \]
Step 7: Convert moles of magnesium into grams.
Molar mass of magnesium is:
\[ 24\;g\;mol^{-1} \] Mass of magnesium required is:
\[ Mass = Moles \times Molar\;mass \] \[ Mass = 0.040 \times 24 \] \[ Mass = 0.96\;g \]
Step 8: Final conclusion.
Therefore, the amount of \(Mg\) required to completely reduce \(100\;ml\) of \(0.1\;M\;NO_3^-\) solution is:
\[ \boxed{0.96\;g} \]
In the reaction, nitrate ion \(NO_3^-\) is reduced to ammonia \(NH_3\).
The oxidation state of nitrogen in \(NO_3^-\) is calculated as:
\[ x+3(-2)=-1 \] \[ x-6=-1 \] \[ x=+5 \] So, nitrogen has oxidation state \(+5\) in \(NO_3^-\).
In \(NH_3\), hydrogen has oxidation state \(+1\). Therefore, oxidation state of nitrogen is:
\[ x+3(+1)=0 \] \[ x+3=0 \] \[ x=-3 \] So, nitrogen has oxidation state \(-3\) in \(NH_3\).
Step 2: Calculate the number of electrons gained by nitrate ion.
Nitrogen changes from \(+5\) to \(-3\).
Therefore, decrease in oxidation state is:
\[ +5-(-3)=8 \] Hence, one \(NO_3^-\) ion gains \(8\) electrons during reduction.
\[ NO_3^- + 8e^- \longrightarrow NH_3 \]
Step 3: Calculate the electrons lost by magnesium.
Magnesium changes from \(Mg\) to \(Mg^{2+}\).
So, magnesium loses \(2\) electrons:
\[ Mg \longrightarrow Mg^{2+}+2e^- \] Thus, \(1\) mole of \(Mg\) gives \(2\) moles of electrons.
Step 4: Find the mole ratio between \(NO_3^-\) and \(Mg\).
One mole of \(NO_3^-\) requires \(8\) moles of electrons.
One mole of \(Mg\) supplies \(2\) moles of electrons.
Therefore, moles of \(Mg\) required for \(1\) mole of \(NO_3^-\) are:
\[ \frac{8}{2}=4 \] So, the balanced electron relation is:
\[ 1\;mole\;NO_3^- \equiv 4\;moles\;Mg \]
Step 5: Calculate moles of nitrate ion present.
Given volume of \(NO_3^-\) solution is:
\[ 100\;ml = 0.100\;L \] Molarity of \(NO_3^-\) solution is:
\[ 0.1\;M \] Moles of \(NO_3^-\) are calculated by:
\[ \text{Moles} = Molarity \times Volume\;in\;litres \] \[ \text{Moles of }NO_3^- = 0.1 \times 0.100 \] \[ =0.010\;mol \]
Step 6: Calculate moles of magnesium required.
Since \(1\) mole of \(NO_3^-\) requires \(4\) moles of \(Mg\),
\(0.010\) mole of \(NO_3^-\) will require:
\[ 0.010 \times 4 = 0.040\;mol \] Thus, required moles of magnesium are:
\[ 0.040\;mol \]
Step 7: Convert moles of magnesium into grams.
Molar mass of magnesium is:
\[ 24\;g\;mol^{-1} \] Mass of magnesium required is:
\[ Mass = Moles \times Molar\;mass \] \[ Mass = 0.040 \times 24 \] \[ Mass = 0.96\;g \]
Step 8: Final conclusion.
Therefore, the amount of \(Mg\) required to completely reduce \(100\;ml\) of \(0.1\;M\;NO_3^-\) solution is:
\[ \boxed{0.96\;g} \]
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