Metallic silver reacts with nitric acid according to the following balanced equation- 3Ags - 4HNO3aq - 3AgNO3aq - NOg - 2H2OlGiven molar masses for Ag- 108gmol and for nitric acid- 63gmol-Determines the minimum volume in milliliters mL of 0-8 M HNO3 needed to dissolve 2-0 g of silver metal-

The correct option is B 2.0× 4× 1000108× 3× 8.0 3Ag (s)+4HNO_3 (aq) ⟶ 3AgM_3(aq)+NO(g)+2H_2O Therefore, 3 moles of Ag requires 4 mol ...

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Standard XII

Chemistry

Ammonia

Metallic silv...

Question

Metallic silver reacts with nitric acid according to the following balanced equation.
$3 A g (s) + 4 H N O_{3} (a q) \to 3 A g N O_{3} (a q) + N O (g) + 2 H_{2} O (l)$
Given molar masses for $A g, 108 \frac{g}{m o l}$ and for nitric acid, $63 \frac{g}{m o l}$ .
Determines the minimum volume in milliliters (mL) of $0.8 M H N O_{3}$ needed to dissolve $2.0 g$ of silver metal.

\frac{2.0 \times 4}{108 \times 3 \times 8.0}

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\frac{2.0 \times 4 \times 1000}{108 \times 3 \times 8.0}

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\frac{4 \times 2.0 \times 8.0}{108 \times 3}

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\frac{2.0 \times 4 \times 8.0}{108 \times 3 \times 1000}

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Solution

The correct option is B $\frac{2.0 \times 4 \times 1000}{108 \times 3 \times 8.0}$
$3 A g (s) + 4 H N O_{3} (a q) ⟶ 3 A g M_{3} (a q) + N O (g) + 2 H_{2} O$
Therefore, $3$ moles of $A g$ requires $4$ moles of $H N O_{3}$ to dissolve.
$2 g m$ of $A g$ metal $\equiv \frac{2}{108} = 0.0185$ moles of $A g$ .
Therefore $0.0185$ moles of $A g$ requires $\frac{4}{3} \times 0.0185$ moles i.e. $\frac{4}{3} \times \frac{2}{108}$ moles of $H N O_{3}$ .
$∴$ The volume of $0.8 M$ $H N O_{3}$ needed will be
$\Rightarrow \frac{4}{3} \times \frac{2}{108} \times \frac{1000}{0.8} = \frac{4}{3} \times \frac{2}{108} \times \frac{1000}{8}$
$∴$ Answer will be Option B.

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Metallic silver reacts with nitric acid according to the following balanced equation.3Ag(s)+4HNO3(aq)→3AgNO3(aq)+NO(g)+2H2O(l)Given molar masses for Ag,108gmol and for nitric acid, 63gmol.Determines the minimum volume in milliliters (mL) of 0.8 M HNO3 needed to dissolve 2.0 g of silver metal.