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Chapter 9: Problem 149

Consider the unbalanced chemical equation \(\mathrm{CaC}_{2}+\mathrm{CO} \rightarrow \mathrm{C}+\mathrm{CaCO}_{3}\) When the reaction is complete, \(135.4 \mathrm{~g}\) of \(\mathrm{CaCO}_{3}\) produced and \(38.5 \mathrm{~g}\) of \(\mathrm{CaC}_{2}\) is left over. Assuming the reaction had a \(100 \%\) yield, what were the mass of the two reactants at the beginning of the reaction

Short Answer

Expert verified
In conclusion, the initial masses of the reactants are: CaC2: 125.3 g CO: 75.8 g

Step by step solution

01

1. Balance the chemical equation

To balance the equation, we need the same number of each type of atom on both sides. We will adjust the coefficients of each compound to achieve this. The balanced equation is: \(CaC_{2} + 2CO \rightarrow C + CaCO_{3}\). Now that we have a balanced chemical equation, we can proceed to calculate the initial masses of each reactant.
02

2. Calculate the initial mass of CaC2

First, we need to determine the initial mass of CaC2 before the reaction. We know the amount produced of CaCO3 (135.4 g) and the amount of unreacted CaC2 (38.5 g). To do this, we first find the molar mass of CaC2 and CaCO3: CaC2: \(Ca (40.08 g/mol) + 2 \times C (12.01 g/mol) = 64.1 g/mol\) CaCO3: \(Ca (40.08 g/mol) + C (12.01 g/mol) + 3 \times O (16.00 g/mol) = 100.09 g/mol\) Then, we will use stoichiometry to calculate the moles of CaC2 that reacted: \[\frac{135.4\, g \, CaCO_3}{100.09\, g \, CaCO_3/mol} \times \frac{1\, mol\, CaC_2}{1\, mol\, CaCO_3} = 1.3534 \,mol\, CaC_2\] Now, we need to find the initial mass of CaC2: Initial moles of CaC2 = moles of reacted CaC2 + moles of leftover CaC2 To convert the leftover CaC2 to moles, divide the given mass by its molar mass: \(\frac{38.5\, g \, CaC_2}{64.1\, g \, CaC_2/mol} = 0.6006\, mol\, CaC_2\) Initial moles of CaC2 = 1.3534 mol + 0.6006 mol = 1.9540 mol To find the initial mass of CaC2, multiply the initial moles of CaC2 by its molar mass: \(1.9540\,mol\, CaC_2 \times 64.1\, g\, CaC_2 / mol = 125.3\,g\, CaC_2\)
03

3. Calculate the initial mass of CO

Now, we will use the stoichiometric ratio from the balanced equation to find the initial mass of CO: \[\frac{1.3534\, mol \, CaC_2}{1\, mol\, CaC_2} \times \frac{2\, mol\, CO}{1\, mol\, CaC_2} = 2.7068\, mol\, CO\] Find the molar mass of CO: CO: \(C (12.01 g/mol) + O (16.00 g/mol) = 28.01 g/mol\) Calculate the initial mass of CO using the moles of CO and its molar mass: \(2.7068\,mol\, CO \times 28.01 \, g \, CO / mol = 75.8 \,g\, CO\) In conclusion, the initial masses of the reactants are: CaC2: 125.3 g CO: 75.8 g

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Key Concepts

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

Chemical Equation Balancing
To understand the importance of balancing chemical equations, let's start with the law of conservation of mass. This fundamental principle of chemistry states that in a chemical reaction, the mass of the reactants must equal the mass of the products. Therefore, balancing a chemical equation ensures that the number of atoms for each element is the same on both sides of the reaction.

In our exercise, we balance the equation \(CaC_{2} + CO \rightarrow C + CaCO_{3}\) by considering each atom. The initial unbalanced equation does not have the same number of carbon and oxygen atoms on both sides. By adding a coefficient of 2 before CO on the reactant side, we balance the carbon and oxygen atoms, resulting in \(CaC_{2} + 2CO \rightarrow C + CaCO_{3}\). This is crucial because it allows us to use stoichiometry to relate the amounts of reactants and products accurately.
Molar Mass Calculation
Molar mass is a fundamental concept in chemistry that refers to the mass of one mole of a substance, typically expressed in grams per mole (g/mol). The molar mass can be calculated by summing the atomic masses of all the atoms in a molecule. Atomic masses can be found on the periodic table and provide the weight of one mole of an element.

For instance, in our exercise, to find the molar mass of calcium carbide (\(CaC_{2}\)), we add the atomic mass of calcium (Ca) to twice the atomic mass of carbon (C) because there are two carbon atoms in the compound. This gives us \(Ca (40.08 g/mol) + 2 \times C (12.01 g/mol) = 64.10 g/mol\). Similarly, for calcium carbonate (\(CaCO_{3}\)), the calculation is \(Ca (40.08 g/mol) + C (12.01 g/mol) + 3 \times O (16.00 g/mol) = 100.09 g/mol\). Knowing these values is essential for converting between mass and moles of a substance.
Reactant Mass Calculation
Using stoichiometry, the quantity of reactants or products in a chemical reaction can be calculated. Stoichiometry hinges on the balanced equation and the molar masses of the substances involved. With this approach, we can find out how much of one reactant is needed to react completely with another, or how much product can be produced from certain amounts of reactants.

In the provided exercise, once we know the molar mass, we can calculate the initial mass of calcium carbide (\(CaC_{2}\)) by adding the mass of the unreacted \(CaC_{2}\) to the equivalent mass of \(CaC_{2}\) that reacted to form \(CaCO_{3}\). Similarly, by utilizing the stoichiometric ratios from the balanced equation, we calculated the initial mass of carbon monoxide (CO) required for the reaction. The accurate determination of the molar masses and understanding the stoichiometric relationships allow us to precisely calculate the reactant masses needed to produce a certain amount of product.

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Most popular questions from this chapter

Silicon nitride, \(\mathrm{Si}_{3} \mathrm{~N}_{4}\), is a ceramic material capable of withstanding high temperatures. It can be produced using the following unbalanced reaction: \(\mathrm{SiCl}_{4}+\mathrm{NH}_{3} \rightarrow \mathrm{Si}_{3} \mathrm{~N}_{4}+\mathrm{HCl}\) If \(64.2 \mathrm{~g}\) of \(\mathrm{SiCl}_{4}\) is reacted with \(20.0 \mathrm{~g}\) of \(\mathrm{NH}_{3}\), how many grams of \(\mathrm{Si}_{3} \mathrm{~N}_{4}\) are produced if the reaction has a \(96.0 \%\) yield?

A \(1.000-g\) sample of a liquid that contains only carbon and hydrogen burns in oxygen to produce \(1.284 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) (a) What are the mass percents of the elements present in this sample? (b) What is the empirical formula for this compound? (c) The molar mass of this compound is determined to be about \(71 \mathrm{~g} / \mathrm{mol}\). What is the molecular formula for this compound? (Hint: When attempting this problem, understand that all of the carbon in the compound burned ends up as \(\mathrm{CO}_{2}\), and all of the hydrogen in the compound burned ends up as \(\mathrm{H}_{2} \mathrm{O}\). Also, there is only 1 mole of \(C\) per mole of \(\mathrm{CO}_{2}\), but there are 2 moles of \(\mathrm{H}\) per mole of \(\mathrm{H}_{2} \mathrm{O} .\) )

Consider the following decomposition reaction in which \(47.20 \mathrm{~g}\) of some compound is decomposed into its elements.

Consider the balanced chemical equation \(2 \mathrm{~A}+\mathrm{B} \rightarrow 2 \mathrm{C}+\mathrm{D}\) When \(8.0 \mathrm{~g}\) of A reacts completely with \(6.0 \mathrm{~g}\) of \(\mathrm{B}\), \(10.0 \mathrm{~g}\) of \(\mathrm{C}\) and \(4.0 \mathrm{~g}\) of \(\mathrm{D}\) are produced. Assuming the yield is \(100 \%\), (a) Which has a greater molar mass, A or C? (b) Which has a greater molar mass, A or B? (c) Which has a greater molar mass, A or D? (d) If the molar mass of \(\mathrm{A}\) is \(24.0 \mathrm{~g} / \mathrm{mol}\), determine the molar mass of \(\mathrm{B}, \mathrm{C}\), and \(\mathrm{D}\).

The flavoring agent vanillin contains carbon, hydrogen, and possibly oxygen. When \(0.450 \mathrm{~g}\) of vanillin is subjected to combustion analysis, the results are \(63.08 \% \mathrm{C}\) and \(5.30 \% \mathrm{H}\). If the molar mass is approximately \(152 \mathrm{~g} / \mathrm{mol}\), what is the molecular formula of vanillin?
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