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Chapter 4: Problem 72

Consider the reaction between silver and nitric acid for which the unbalanced equation is $$ \mathrm{Ag}(s)+\mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{Ag}^{+}(a q)+\mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O} $$ (a) Balance the equation. (b) If \(42.50 \mathrm{~mL}\) of \(12.0 \mathrm{M}\) nitric acid furnishes enough \(\mathrm{H}^{+}\) to react with silver, how many grams of silver react?

Short Answer

Expert verified
Answer: 41.24 g of silver reacts with the given nitric acid solution.

Step by step solution

01

Balance the chemical equation

To balance the chemical equation, we need to make sure that the number of atoms of each element remains the same in the reactants and products. The balanced chemical equation is: $$ 3 \mathrm{Ag}(s) + 4 \mathrm{H}^{+}(aq) + 4 \mathrm{NO}_{3}^{-}(aq) \longrightarrow 3 \mathrm{Ag}^{+}(aq) + 2 \mathrm{NO}_{2}(g) + 2 \mathrm{H}_{2}\mathrm{O}(l) $$
02

Find the moles of nitric acid

We are given that the volume of nitric acid is \(42.50 \mathrm{~mL}\) and its molarity is \(12.0 \mathrm{M}\), which means there are \(12.0\) moles of nitric acid in \(1\) liter of the solution. To find the moles of nitric acid, we use the formula: moles of solute = molarity \(\times\) volume (in liters). So, moles of nitric acid \(= 12.0 \mathrm{M} \times 42.50 \times 10^{-3} \mathrm{L} = 0.510 \mathrm{~mol}\).
03

Find the moles of silver that reacted

Using the balanced chemical equation, we can see that \(4\) moles of nitric acid react with \(3\) moles of silver. Using the stoichiometry, we can now set up the ratio: $$ \frac{3 \mathrm{~moles~Ag}}{4 \mathrm{~moles~HNO_3}} = \frac{x \mathrm{~moles~Ag}}{0.510 \mathrm{~moles~HNO_3}} $$ Solving for \(x\) moles of silver: $$ x = \frac{3\cdot 0.510}{4} = 0.3825 \mathrm{~moles~Ag} $$
04

Convert moles of silver to mass of silver

To find the mass of silver that reacted, we multiply the number of moles of silver found in Step 3 by the molar mass of silver: $$ \text{mass of silver} = 0.3825 \mathrm{~moles~Ag} \times 107.87 \frac{\mathrm{g}}{\mathrm{mol}} = 41.24 \mathrm{~g} $$ Therefore, \(41.24 \mathrm{~g}\) of silver react with the given nitric acid.

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Most popular questions from this chapter

Write balanced equations for the following reactions in basic solution. (a) \(\mathrm{Ni}(\mathrm{OH})_{2}(s)+\mathrm{N}_{2} \mathrm{H}_{4}(a q) \longrightarrow \mathrm{Ni}(s)+\mathrm{N}_{2}(g)\) (b) \(\mathrm{Fe}(\mathrm{OH})_{3}(s)+\mathrm{Cr}^{3+}(a q) \longrightarrow \mathrm{Fe}(\mathrm{OH})_{2}(s)+\mathrm{CrO}_{4}{ }^{2-}(a q)\) (c) \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{BrO}_{3}^{-}(a q) \longrightarrow \mathrm{MnO}_{2}(s)+\mathrm{BrO}_{4}^{-}(a q)\) (d) \(\mathrm{H}_{2} \mathrm{O}_{2}(a q)+\mathrm{IO}_{4}^{-}(a q) \longrightarrow \mathrm{IO}_{2}^{-}(a q)+\mathrm{O}_{2}(g)\)

Write a balanced net ionic equation for each of the following acidbase reactions in water. (a) nitrous acid and barium hydroxide (b) potassium hydroxide and hydrofluoric acid (c) aniline \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\right)\) and perchloric acid

What is the molarity of each ion present in aqueous solutions prepared by dissolving \(15.0 \mathrm{~g}\) of the following compounds in water to make \(655 \mathrm{~mL}\) of solution? (a) scandium(III) iodide (b) sodium carbonate (c) magnesium phosphate (d) potassium oxide

Write balanced net ionic equations for the following reactions in acid solution. (a) Liquid hydrazine reacts with an aqueous solution of sodium bromate. Nitrogen gas and bromide ions are formed. (b) Solid phosphorus \(\left(P_{4}\right)\) reacts with an aqueous solution of nitrate to form nitrogen oxide gas and dihydrogen phosphate \(\left(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\right)\) ions. (c) Aqueous solutions of potassium sulfite and potassium permanganate react. Sulfate and manganese(II) ions are formed.

What is the molarity of each ion present in aqueous solutions prepared by dissolving \(20.00 \mathrm{~g}\) of the following compounds in water to make 4.50 L of solution? (a) cobalt(III) chloride (b) nickel(III) sulfate (c) sodium permanganate (d) iron(II) bromide
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