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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 A compound has been isolated that can have either of two possible formulas: (a) \(\mathrm{K}\left[\mathrm{Fe}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{2}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2}\right]\) or (b) \(\mathrm{K}_{3}\left[\mathrm{Fe}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right] .\) To find which is correct, you dissolve a weighed sample of the compound in acid, forming oxalic acid, \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\). You then titrate this acid with potassium permanganate, \(\mathrm{KMnO}_{4}\) (the source of the \(\left.\mathrm{MnO}_{4}-\text { ion }\right) .\) The balanced, net ionic equation for the titration is $$ \begin{aligned} 5 \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(\mathrm{aq})+& 2 \mathrm{MnO}_{4}-(\mathrm{aq})+6 \mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq}) \rightarrow \\ & 2 \mathrm{Mn}^{2+}(\mathrm{aq})+10 \mathrm{CO}_{2}(\mathrm{g})+14 \mathrm{H}_{2} \mathrm{O}(\ell) \end{aligned} $$ Titration of \(1.356 \mathrm{g}\) of the compound requires \(34.50 \mathrm{mL}\) of \(0.108 \mathrm{M} \mathrm{KMnO}_{4} .\) Which is the correct formula of the iron-containing compound: (a) or (b)?

Sodium thiosulfate, \(\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3},\) is used as a "fixer" in black-and-white photography. Suppose you have a botule of sodium thiosulfate and want to determine its purity. The thiosulfate ion can be oxidized with \(\mathrm{I}_{2}\) according to the balanced, net ionic equation $$ \mathrm{I}_{2}(\mathrm{aq})+2 \mathrm{S}_{2} \mathrm{O}_{3}^{2-}(\mathrm{aq}) \rightarrow 2 \mathrm{I}^{-}(\mathrm{aq})+\mathrm{S}_{4} \mathrm{O}_{6}^{2-}(\mathrm{aq}) $$ If you use \(40.21 \mathrm{mL}\) of \(0.246 \mathrm{M} \mathrm{I}_{2}\) in a titration, what is the weight percent of \(\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}\) in a \(3.232-\mathrm{g}\) sample of impure material?

Nickel forms a compound with carbon monoxide, \(\mathrm{Ni}_{x}(\mathrm{CO})_{\gamma}\). To determine its formula, you carefully heat a \(0.0973-\mathrm{g}\) sample in air to convert the nickel to \(0.0426 \mathrm{g}\) of \(\mathrm{NiO}\) and the \(\mathrm{CO}\) to \(0.100 \mathrm{g}\) of \(\mathrm{CO}_{2} .\) What is the empirical formula of \(\mathrm{Ni}_{x}(\mathrm{CO}), 2\)

A Sulfuric acid is listed in a catalog with a concentration of \(95-98 \% .\) A bottle of the acid in the stockroom states that 1.00 I. has a mass of \(1.84 \mathrm{kg} .\) To determine the concentration of sulfuric acid in the stockroom bottle, a student dilutes \(5.00 \mathrm{mL}\) to \(500 .\) mL. She then takes four \(10.00-\mathrm{mL}\). samples and titrates each with standardized sodium hydroxide \((c=0.1760 \mathrm{M}).\) \(\begin{array}{lcccc}\text { Sample } & 1 & 2 & 3 & 4 \\ \text { Volume NaOH (mL) } & 20.15 & 21.30 & 20.40 & 20.35\end{array}\) (a) What is the average concentration of the diluted sulfuric acid sample? (b) What is the mass percent of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) in the original bottle of the acid?

An unknown compound has the formula \(\mathrm{C}_{x} \mathrm{H}, \mathrm{O}_{z} .\) You burn \(0.1523 \mathrm{g}\) of the compound and isolate \(0.3718 \mathrm{g}\) of \(\mathrm{CO}_{2}\) and \(0.1522 \mathrm{g}\) of \(\mathrm{H}_{2} \mathrm{O} .\) What is the empirical formula of the compound? If the molar mass is \(72.1 \mathrm{g} / \mathrm{mol},\) what is the molecular formula?

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