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Chapter 9: Problem 90

Hydrazine, \(\mathrm{N}_{2} \mathrm{H}_{4}\), cmits a large quantity of energy when it reacts with oxygen, which has led to hydrazine's use as a fuel for rockets: $$ \mathrm{N}_{2} \mathrm{H}_{4}(I)+\mathrm{O}_{2}(g) \rightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) $$ How many moles of each of the gaseous products are produced when \(20.0 \mathrm{~g}\) of pure hydrazine is ignited in the presence of \(20.0 \mathrm{~g}\) of pure oxygen? How many grams of each product are produced?

Short Answer

Expert verified
When 20.0 grams of hydrazine is ignited in the presence of 20.0 grams of oxygen, \(0.624\ \text{mol}\) of nitrogen and \(1.248\ \text{mol}\) of water are produced. This corresponds to 17.5 grams of nitrogen and 22.5 grams of water.

Step by step solution

01

Balanced chemical equation

We are given the balanced chemical equation: \(\mathrm{N_{2}H_{4}(l) + O_{2}(g) \rightarrow N_{2}(g) + 2H_{2}O(g)}\)
02

Calculate the number of moles of reactants

First, we need to find the molar masses of hydrazine \(\mathrm{(N_{2}H_{4})}\) and oxygen \(\mathrm{(O_{2})}\). \[\text{Molar mass of N}_{2}\text{H}_{4} = 2\times 14.01\ \text{g/mol (N)} + 4\times 1.008\ \text{g/mol (H)} = 32.05\ \text{g/mol}\] \[\text{Molar mass of O}_{2} = 2\times 16.00\ \text{g/mol} = 32.00\ \text{g/mol}\] Now we can calculate the moles of each reactant: \[n(\mathrm{N_{2}H_{4}}) = \frac{20.0\ \text{g}}{32.05\ \text{g/mol}} = 0.624\ \text{mol}\] \[n(\mathrm{O}_{2}) = \frac{20.0\ \text{g}}{32.00\ \text{g/mol}} = 0.625\ \text{mol}\]
03

Determine the limiting reactant

To identify the limiting reactant, we need to look at the balanced equation and find the reactant with the lower mole ratio. \[n\left(\frac{\text {N}_{2}\text {H}_{4}}{\text{O}_{2}}\right)=\frac{n(\text{N}_{2}\text {H}_{4})}{n(\text{O}_{2})}=\frac{0.624}{0.625}=0.998\] Since the mole ratio of \(\text{N}_{2}\text {H}_{4}\) to \(\text{O}_{2}\) is less than 1, \(\text{N}_{2}\text {H}_{4}\) is the limiting reactant.
04

Use stoichiometry to find the number of moles of each product

Using the balanced equation and the moles of the limiting reactant, we can find the moles of each product: \[n(\text{N}_{2}) = n(\text{N}_{2}\text {H}_{4}) = 0.624\ \text{mol}\] \[n(\text{H}_{2}\text{O}) = 2\times n(\text{N}_{2}\text {H}_{4}) = 2\times 0.624\ \text{mol} = 1.248\ \text{mol}\]
05

Convert the moles of products to grams

Finally, convert the moles of each product to grams using their molar masses: \[\text{Mass of N}_{2} = n(\text{N}_{2}) \times \text{molar mass of N}_{2} = 0.624\ \text{mol} \times 28.02\ \text{g/mol} = 17.5\ \text{g}\] \[\text{Mass of H}_{2}\text{O} = n(\text{H}_{2}\text{O}) \times \text{molar mass of H}_{2}\text{O} = 1.248\ \text{mol} \times 18.02\ \text{g/mol} = 22.5\ \text{g}\] Therefore, when 20.0 grams of hydrazine is ignited in the presence of 20.0 grams of oxygen, 17.5 grams of nitrogen and 22.5 grams of water are produced.

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Key Concepts

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

Chemical Equations
Chemical equations are the backbone of understanding chemical reactions. They represent what occurs at the atomic level during a reaction. In chemical equations, reactants (starting substances) and products (substances formed) are separated by an arrow. For example, in the equation \(\mathrm{N_{2}H_{4}(l) + O_{2}(g) \rightarrow N_{2}(g) + 2H_{2}O(g)}\), Hydrazine \((\mathrm{N_2H_4})\) and oxygen \((\mathrm{O_2})\) are the reactants, while nitrogen \((\mathrm{N_2})\) and water \((\mathrm{H_2O})\) are the products.
Balancing chemical equations is crucial to reflect the law of conservation of mass, which states that mass is neither created nor destroyed in a chemical reaction. This means the number of atoms for each element involved in the reaction must be the same on both sides of the equation. In our example, we ensure that there are equal numbers of nitrogen, hydrogen, and oxygen atoms on both sides of the equation. This step helps correctly predict the quantities of reactants and products involved in a reaction.
Limiting Reactant
In stoichiometry, the limiting reactant is the substance that is entirely consumed first in a chemical reaction, limiting the amount of product that can be formed. To find the limiting reactant, compare the mole ratio of the reactants you have with the mole ratio in the balanced equation.
In the given problem, both hydrazine and oxygen are used as reactants. Start by calculating their moles: hydrazine has \(0.624\ \text{mol}\) and oxygen has \(0.625\ \text{mol}\). Look at the balanced chemical equation: it shows a 1:1 ratio of hydrazine to oxygen. Thus, theoretically, we would need the same amount of moles for complete reaction. However, since \(0.624\) moles of hydrazine are slightly less than \(0.625\) moles of oxygen, hydrazine becomes the limiting reactant.
This concept ensures that no extra calculations are made for producing products from an amount of reactants that do not exist, making it a vital skill for predicting chemical yields accurately.
Molar Mass Calculation
Molar mass is essential for converting grams of a substance to moles. It acts as a bridge in stoichiometric calculations between mass and number of moles, allowing you to perform reaction predictions correctly.
To calculate the molar mass of a compound, sum the atomic masses (from the periodic table) of all atoms in a molecule. For example, hydrazine \((\mathrm{N_2H_4})\) has two nitrogen atoms (each \(14.01\ \text{g/mol}\)) and four hydrogen atoms (each \(1.008\ \text{g/mol}\)) leading to a molar mass of \(32.05\ \text{g/mol}\). Similarly, oxygen \((\mathrm{O_2})\)'s molar mass is \(32.00\ \text{g/mol}\) as it consists of two oxygen atoms (each \(16.00\ \text{g/mol}\)).
Understanding these simple conversions is crucial for calculating how many moles of reactants will react and the moles of products formed, completely from the given mass of substances.

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Most popular questions from this chapter

For cach of the following balanced chemical cquations, calculate how many moles and how many grams of each product would be produced by the complete conversion of 0.50 mole of the reactant indicated in boldface. State clearly the mole ratio used for each conversion. a. \(\mathrm{NH}_{3}(g)+\mathrm{HCl}(g) \rightarrow \mathrm{NH}_{4} \mathrm{Cl}(s)\) b. \(\mathrm{CH}_{4}(g)+4 \mathrm{~S}(s) \rightarrow \mathrm{CS}_{2}(l)+2 \mathrm{H}_{2} \mathrm{~S}(g)\) c. \(\mathbf{P C l}_{3}(l)+3 \mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{H}_{3} \mathrm{PO}_{3}(a q)+3 \mathrm{HCl}(a q)\) d. \(\mathrm{NaOH}(s)+\mathrm{CO}_{2}(g) \rightarrow \mathrm{NaHCO}_{3}(s)\)

What do the coefficients of a balanced chemical equation tell us about the proportions in which atoms and molecules react on an individual (microscopic) basis?

Barium chloride solutions are used in chemical analysis for the quantitative precipitation of sulfate ion from solution. $$ \mathrm{Ba}^{2+}(a q)+\mathrm{SO}_{4}^{2-}(a q) \rightarrow \mathrm{BaSO}_{4}(s) $$ Suppose a solution is known to contain on the order of \(150 \mathrm{mg}\) of sulfate ion. What mass of barium chloride should be added to guarantee precipitation of all the sulfate ion?

The text explains that one reason why the actual yicld for a reaction may be less than the theoretical yicld is side reactions. Suggest some other reasons why the percent yield for a reaction might not be \(100 \%\).

Natural waters often contain relatively high levels of calcium ion, \(\mathrm{Ca}^{2+},\) and hydrogen carbonate ion (bicarbonate), \(\mathrm{HCO}_{3}^{-}\), from the leaching of minerals into the water. When such water is used commercially or in the home, heating of the water leads to the formation of solid calcium carbonate, \(\mathrm{CaCO}_{3}\), which forms a deposit ("scale") on the interior of boilers, pipes, and other plumbing fixtures. $$ \mathrm{Ca}\left(\mathrm{HCO}_{3}\right)_{2}(a q) \rightarrow \mathrm{CaCO}_{3}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) $$ If a sample of well water contains \(2.0 \times 10^{-3} \mathrm{mg}\) of \(\mathrm{Ca}\left(\mathrm{HCO}_{3}\right)_{2}\) per milliliter, what mass of \(\mathrm{CaCO}_{3}\) scale would \(1.0 \mathrm{~mL}\) of this water be capable of depositing?
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