/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 62 Hydrazine, \(\mathrm{N}_{2} \mat... [FREE SOLUTION] | 91Ó°ÊÓ

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Hydrazine, \(\mathrm{N}_{2} \mathrm{H}_{4},\) a base like ammonia, can react with sulfuric acid. \(2 \mathrm{N}_{2} \mathrm{H}_{4}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq}) \rightarrow 2 \mathrm{N}_{2} \mathrm{H}_{5}+(\mathrm{aq})+\mathrm{SO}_{4}^{2-}(\mathrm{aq})\) What mass of hydrazine reacts with \(250 .\) mL. of \(0.146 \mathrm{M}\) \(\mathrm{H}_{2} \mathrm{SO}_{4} ?\)

Short Answer

Expert verified
2.34 g of hydrazine reacts with 250 mL of 0.146 M \(\mathrm{H}_2 \mathrm{SO}_4\).

Step by step solution

01

Calculate moles of sulfuric acid

First, find the moles of sulfuric acid (\( \mathrm{H}_2 \mathrm{SO}_4 \)) using the volume and molarity given. The formula to use is: \( \text{moles} = \text{molarity} \times \text{volume in liters} \). Convert the volume from mL to L: \[250 \text{ mL} = 0.250 \text{ L}\]Then calculate the moles:\[\text{moles} = 0.146 \text{ M} \times 0.250 \text{ L} = 0.0365 \text{ moles of } \mathrm{H}_2 \mathrm{SO}_4\]
02

Use stoichiometry to find moles of hydrazine

From the balanced equation, the molar ratio of \( \mathrm{N}_2 \mathrm{H}_4 \) to \( \mathrm{H}_2 \mathrm{SO}_4 \) is 2:1. Therefore, the moles of hydrazine is twice the moles of sulfuric acid. Calculate the moles of \( \mathrm{N}_2 \mathrm{H}_4 \):\[\text{moles of } \mathrm{N}_2 \mathrm{H}_4 = 2 \times 0.0365 = 0.0730 \text{ moles}\]
03

Calculate the mass of hydrazine

Now, calculate the mass of \( \mathrm{N}_2 \mathrm{H}_4 \) using its molar mass. The molar mass of \( \mathrm{N}_2 \mathrm{H}_4 \) is calculated by adding the atomic masses: \[ (2 \times 14.01 \text{ g/mol for N}) + (4 \times 1.01 \text{ g/mol for H}) = 32.05 \text{ g/mol}\]Finally, calculate the mass:\[\text{mass} = 0.0730 \text{ moles} \times 32.05 \text{ g/mol} = 2.338 \text{ g}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molarity in Chemical Reactions
Molarity is a fundamental concept in chemistry that refers to the concentration of a solution, specifically the number of moles of a solute per liter of solution. This is crucial when dealing with chemical reactions in aqueous solutions, as it allows for precise control over how much of a substance is involved in a reaction. In this exercise, the molarity of sulfuric acid, denoted as \[0.146 \, \text{M}\], indicates that there are 0.146 moles of sulfuric acid present in every liter of solution.
  • To calculate the amount of a chemical needed or produced, we start by using the molarity equation: \[ \text{moles} = \text{molarity} \times \text{volume (L)} \]
  • This implies first converting all volume measurements to liters as molarity is defined per liter.
In the original problem, we had 250 mL of sulfuric acid, which converts to 0.250 L. The number of moles of \( \text{H}_2 \text{SO}_4 \) is found by multiplying the molarity \((0.146 \, \text{M})\) by the volume \((0.250 \, \text{L})\), yielding 0.0365 moles. This step is crucial as it forms the basis for further mole calculations.
Understanding Chemical Reactions and Mole Calculations
Chemical reactions involve rearrangements of molecules, and the balanced equation provides a ratio showing how reactants transform into products. For the reaction between hydrazine \( \text{N}_2 \text{H}_4 \) and sulfuric acid \( \text{H}_2 \text{SO}_4 \), the balanced equation given is:\[ 2 \text{N}_2 \text{H}_4(\text{aq}) + \text{H}_2 \text{SO}_4(\text{aq}) \rightarrow 2 \text{N}_2 \text{H}_5^+(\text{aq}) + \text{SO}_4^{2-}(\text{aq}) \]
  • This equation tells us for every mole of sulfuric acid, two moles of hydrazine are required.
  • The stoichiometric coefficients (the numbers in front of the molecules) are key to understanding how much of each reactant is needed or how much product will be formed.
From our molarity calculation, we found 0.0365 moles of sulfuric acid available. Using the stoichiometric ratios from the balanced equation, we multiply by 2 (from the 2:1 ratio) to find that 0.0730 moles of hydrazine are needed to fully react with the given sulfuric acid.
Calculating the Mass from Moles
Once we ascertain the number of moles needed for a reaction, the next step often involves converting these moles into a mass for practical purposes. In our problem, we calculated that 0.0730 moles of hydrazine \( \text{N}_2 \text{H}_4 \) are necessary. To convert moles to mass, we use the molar mass, which is the weight of one mole of a given substance.
  • The molar mass of hydrazine is determined by adding the atomic masses of its constituent atoms.
  • For hydrazine: - \[ 2 \times 14.01 \, \text{g/mol for N} + 4 \times 1.01 \, \text{g/mol for H} = 32.05 \, \text{g/mol} \]
Using the molar mass, the relationship \( \text{mass} = \text{moles} \times \text{molar mass} \) allows us to calculate: \[ 0.0730 \, \text{moles} \times 32.05 \, \text{g/mol} = 2.338 \, \text{g} \]Thus, 2.338 grams of hydrazine are necessary to react completely with the sulfuric acid provided.

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Most popular questions from this chapter

A Cloth can be waterproofed by coating it with a silicone layer. This is done by exposing the cloth to \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{SiCl}_{2}\) vapor. The silicon compound reacts with OH groups on the cloth to form a waterproofing film (density \(=1.0 \mathrm{g} / \mathrm{cm}^{3}\) ) of \(\left[\left(\mathrm{CH}_{3}\right)_{2} \mathrm{SiO}\right]_{m},\) where \(n\) is a large integer number. \(n\left(\mathrm{CH}_{3}\right)_{2} \mathrm{SiCl}_{2}+2 n \mathrm{OH}^{-} \rightarrow\) $$ 2 n \mathrm{Cl}^{-}+n \mathrm{H}_{2} \mathrm{O}+\left[\left(\mathrm{CH}_{3}\right)_{2} \mathrm{SiO}\right]_{n} $$ The coating is added layer by layer, with each layer of \(\left[\left(\mathrm{CH}_{3}\right)_{2} \mathrm{SiO}\right]_{n}\) being \(0.60 \mathrm{nm}\) thick. Suppose you want to waterproof a piece of cloth that is 3.00 square meters, and you want 250 layers of waterproofing compound on the cloth. What mass of \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{SiCl}_{2}\) do you need?

An Alka-Seltzer tablet contains exactly \(100 .\) mg of citric acid, \(\mathrm{H}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7},\) plus some sodium bicarbonate. What mass of sodium bicarbonate is required to consume \(100 .\) mg of citric acid by the following reaction? \(\mathrm{H}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}(\mathrm{aq})+3 \mathrm{NaHCO}_{3}(\mathrm{aq}) \rightarrow\) $$ 3 \mathrm{H}_{2} \mathrm{O}(\ell)+3 \mathrm{CO}_{2}(\mathrm{g})+\mathrm{Na}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}(\mathrm{aq}) $$

A Commercial sodium "hydrosulfite" is \(90.1 \%\) \(\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{4} .\) The sequence of reactions used to prepare the compound is $$ \mathrm{Zn}(\mathrm{s})+2 \mathrm{SO}_{2}(\mathrm{g}) \rightarrow \mathrm{ZnS}_{2} \mathrm{O}_{4}(\mathrm{s}) $$ \(\mathrm{ZnS}_{2} \mathrm{O}_{4}(\mathrm{s})+\mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{aq}) \rightarrow \mathrm{ZnCO}_{3}(\mathrm{s})+\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{4}(\mathrm{aq})\) (a) What mass of pure \(\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{4}\) can be prepared from \(125 \mathrm{kg}\) of \(\mathrm{Zn}, 500 . \mathrm{g}\) of \(\mathrm{SO}_{2},\) and an excess of \(\mathrm{Na}_{2} \mathrm{CO}_{3} ?\) (b) What mass of the commercial product would contain the \(\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{4}\) produced using the amounts of reactants in part (a)?

Anhydrous calcium chloride is a good drying agent because it will rapidly pick up water. Suppose you have stored some carefully dried \(\mathrm{CaCl}_{2}\) in a desiccator. Unfortunately, someone did not close the top of the desiccator tightly, and the \(\mathrm{CaCl}_{2}\) became partially hydrated. A \(150-\mathrm{g}\) sample of this partially hydrated material was dissolved in \(80 \mathrm{g}\) of hot water. When the solution was cooled to \(20^{\circ} \mathrm{C}, 74.9 \mathrm{g}\) of \(\mathrm{CaCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}\) precipitated. Knowing the solubility of calcium chloride in water at \(20^{\circ} \mathrm{C}\) is \(74.5 \mathrm{g} \mathrm{CaCl}_{2} / 100 \mathrm{g}\) water, determine the water content of the \(150-\mathrm{g}\) sample of partially hydrated calcium chloride (in moles of water per mole of \(\mathrm{CaCl}_{2}\) ).

What mass of \(\mathrm{Na}_{2} \mathrm{CO}_{3},\) in grams, is required for complete reaction with \(50.0 \mathrm{mL}\) of \(0.125 \mathrm{M} \mathrm{HNO}_{3} ?\) \(\mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{aq})+2 \mathrm{HNO}_{3}(\mathrm{aq}) \rightarrow\) $$ 2 \mathrm{NaNO}_{3}(\mathrm{aq})+\mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\ell) $$

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