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Chapter 9: Problem 29

When elemental carbon is burned in the open atmosphere, with plenty of oxygen gas prescnt, the product is carbon dioxide. $$ \mathrm{C}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g) $$ However, when the amount of oxygen present during the burning of the carbon is restricted, carbon monoxide is more likely to result. $$ 2 \mathrm{C}(s)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{CO}(g) $$ What mass of each product is expected when a \(5.00-\mathrm{g}\) sample of pure carbon is burned under cach of these conditions?

Short Answer

Expert verified
When a 5.00g sample of pure carbon is burned, we can expect 18.3g of CO鈧� to be produced in the presence of plenty of oxygen, and 11.6g of CO to be produced when the amount of oxygen is restricted.

Step by step solution

01

Convert the mass of carbon to moles.

Using the molar mass of carbon (C), 12.01g/mol, we can convert the 5.00g sample of pure carbon to moles: moles of C = (mass of C) / (molar mass of C) moles of C = \( \frac{5.00g}{12.01g/mol} \)= 0.416 moles
02

Use stoichiometry to find moles of each product.

For each of the conditions, we can relate the moles of carbon reacted to the moles of products formed using stoichiometry. 1. With plenty of oxygen (forming CO鈧�): According to the balanced equation: C(s) + O鈧�(g) 鈫� CO鈧�(g) 1 mole of C forms 1 mole of CO鈧� So, moles of CO鈧� formed = moles of C = 0.416 moles 2. With restricted oxygen (forming CO): According to the balanced equation: 2C(s) + O鈧�(g) 鈫� 2CO(g) 2 moles of C form 2 moles of CO So, moles of CO formed = moles of C = 0.416 moles
03

Convert moles of products to mass.

Now that we have the moles of each product, we can convert them to mass using the molar mass of the products: 1. Mass of CO鈧� formed: molar mass of CO鈧� = 12.01g/mol (C) + 2 脳 16.00g/mol (O) = 44.01g/mol mass of CO鈧� = moles of CO鈧� 脳 molar mass of CO鈧� mass of CO鈧� = 0.416 moles 脳 44.01g/mol = 18.3g 2. Mass of CO formed: molar mass of CO = 12.01g/mol (C) + 16.00g/mol (O) = 28.01g/mol mass of CO = moles of CO 脳 molar mass of CO mass of CO = 0.416 moles 脳 28.01g/mol = 11.6g
04

Summarize the results.

When a 5.00g sample of pure carbon is burned under the two different conditions, we expect the following masses of products: 1. With plenty of oxygen: 18.3g of CO鈧� 2. With restricted oxygen: 11.6g of CO

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Key Concepts

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

Chemical Reactions
Chemical reactions are processes where reactants transform into products. This change involves making and breaking chemical bonds, leading to new substances with different properties. For example, when carbon burns in oxygen, it primarily produces carbon dioxide (CO2), but in a limited oxygen environment, carbon monoxide (CO) can form instead. Understanding stoichiometry is essential for predicting the amounts of products formed in these reactions.

In the given exercise, we look at two reactions:
  • Burning carbon in excess oxygen: \( \mathrm{C}(s) + \mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g) \)
  • Burning carbon in restricted oxygen: \( 2\mathrm{C}(s) + \mathrm{O}_{2}(g) \rightarrow 2\mathrm{CO}(g) \)
The process of burning, in this context, is a combustion reaction, which is an exothermic process releasing energy, often as heat.
Molar Mass
The molar mass of an element or compound is the mass of one mole of that substance. It is an essential concept in chemistry because it allows us to convert between mass and moles, which is a key step in stoichiometry. The molar mass is expressed in units of grams per mole (g/mol).

For instance, carbon has a molar mass of 12.01 g/mol. A mole is Avogadro's number (approximately \(6.02 \times 10^{23}\)) of atoms. By knowing the molar mass, we can calculate how many moles are in a given sample. In the example exercise, we begin by converting 5.00 grams of carbon to moles using its molar mass.
Chemical Equations
Chemical equations are symbolic representations of chemical reactions, where the reactant molecules are shown on the left and the products on the right. They must be balanced, meaning the number of atoms for each element must be the same on both sides of the equation.

A balanced chemical equation ensures the law of conservation of mass is upheld. It also provides the stoichiometric coefficients that tell us the ratios of reactants to products. These ratios are fundamental in determining how much product can be formed from given amounts of reactants, as demonstrated in the exercise problem.
Conversion of Moles to Grams
Conversion of moles to grams is a common task in chemistry, and it relies on the molar mass of a substance. Once you have determined the number of moles of a reactant or product, you can then multiply by its molar mass to find the equivalent mass in grams.

In our exercise, after finding the number of moles of carbon dioxide and carbon monoxide produced, the final step is to convert these mole quantities into masses using the molar masses of CO2 and CO. This process is key to predicting and measuring the actual mass of product that results from a chemical reaction.

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Most popular questions from this chapter

For each of the following unbalanced chemical equations, suppose that exactly \(1.00 \mathrm{~g}\) of each reactant is taken. Determine which reactant is limiting, and calculate what mass of the product in boldface is expected (assuming that the limiting reactant is completely consumed). a. \(\mathrm{CS}_{2}(l)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+\mathrm{SO}_{2}(g)\) b. \(\mathrm{NH}_{3}(g)+\mathrm{CO}_{2}(g) \rightarrow \mathrm{CN}_{2} \mathrm{H}_{4} \mathrm{O}(s)+\mathrm{H}_{2} \mathrm{O}(g)\) c. \(\mathrm{H}_{2}(g)+\mathrm{MnO}_{2}(s) \rightarrow \mathrm{MnO}(s)+\mathbf{H}_{2} \mathrm{O}(g)\) d. \(\mathrm{I}_{2}(l)+\mathrm{Cl}_{2}(g) \rightarrow \mathbf{I C l}(g)\)

If baking soda (sodium hydrogen carbonate) is heated strongly, the following reaction occurs: $$ 2 \mathrm{NaHCO}_{3}(s) \rightarrow \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}_{2}(g) $$ Calculate the mass of sodium carbonate that will remain if a 1.52 - \(\mathrm{g}\) sample of sodium hydrogen carbonate is heated.

Consider the following unbalanced chemical equation for the combustion of pentane \(\left(\mathrm{C}_{5} \mathrm{H}_{12}\right)\) $$ \mathrm{C}_{5} \mathrm{H}_{12}(l)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) $$ If a 20.4 -gram sample of pentane is burned in excess oxygen, what mass of water can be produced, assuming \(100 \%\) yield?

What does it mean to say that the balanced chemical equation for a reaction describes the stoichiometry of the reaction?

Solid copper can be produced by passing gaseous ammonia over solid copper(II) oxide at high temperatures. The other products of the reaction are nitrogen gas and water vapor. The balanced equation for this reaction is: $$ 2 \mathrm{NH}_{3}(g)+3 \mathrm{CuO}(s) \rightarrow \mathrm{N}_{2}(g)+3 \mathrm{Cu}(s)+3 \mathrm{H}_{2} \mathrm{O}(g) $$ What is the theoretical yield of solid copper that should form when \(18.1 \mathrm{~g}\) of \(\mathrm{NH}_{3}\) is reacted with \(90.4 \mathrm{~g}\) of \(\mathrm{CuO}\) ? If only \(45.3 \mathrm{~g}\) of copper is actually collected, what is the percent yield?
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