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Thionyl chloride, \(\mathrm{SOCl}_{2}\), is used as a very powerful drying agent in many synthetic chemistry experiments in which the presence of even small amounts of water would be detrimental. The unbalanced chemical equation is $$ \mathrm{SOCl}_{2}(l)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{SO}_{2}(g)+\mathrm{HCl}(g) $$ Calculate the mass of water consumed by complete reaction of \(35.0 \mathrm{~g}\) of \(\mathrm{SOCl}_{2}\)

Short Answer

Expert verified
The mass of water consumed by the complete reaction of 35.0 g of SOClâ‚‚ is 5.89 g.

Step by step solution

01

Balance the chemical equation

To balance the chemical equation, we need to make sure the number of atoms of each element is the same on both sides of the equation. For the given equation: $$ \mathrm{SOCl}_{2}(l)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{SO}_{2}(g)+\mathrm{HCl}(g) $$ We can see that there are the same number of S and O atoms on both sides. However, there are 2 H atoms and 2 Cl atoms on the left and only 1 H atom and 1 Cl atom on the right. To balance this equation, we can place a coefficient of 2 in front of HCl on the right side: $$ \mathrm{SOCl}_{2}(l)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{SO}_{2}(g)+2\mathrm{HCl}(g) $$ Now, the chemical equation is balanced.
02

Calculate the number of moles of SOClâ‚‚

Now that we have the balanced chemical equation, we can use the given mass of SOCl₂ (35.0 g) to calculate the number of moles present. We'll use the molar mass of SOCl₂ which is: Molar mass of S = 32.06 g/mol Molar mass of O = 16.00 g/mol Molar mass of Cl = 35.45 g/mol Molar mass of SOCl₂: 32.06 + 16.00 + (2 × 35.45) = 106.92 g/mol Number of moles of SOCl₂ = (35.0 g) / (106.92 g/mol) = 0.327 mol
03

Determine the number of moles of Hâ‚‚O consumed

According to the balanced chemical equation, 1 mole of SOClâ‚‚ reacts with 1 mole of Hâ‚‚O. Therefore, the number of moles of Hâ‚‚O consumed is equal to the number of moles of SOClâ‚‚. Number of moles of Hâ‚‚O consumed = 0.327 mol
04

Calculate the mass of Hâ‚‚O consumed

To find the mass of H₂O consumed, we can multiply the number of moles by its molar mass (Molar mass of H₂O = 18.02 g/mol). Mass of H₂O consumed = (0.327 mol) × (18.02 g/mol) = 5.89 g Thus, the mass of water consumed by the complete reaction of 35.0 g of SOCl₂ is 5.89 g.

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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 chemistry. They represent chemical reactions by using symbols for the elements and formulas for the compounds. In our example of a reaction involving thionyl chloride (\(\mathrm{SOCl}_2\)) and water (\(\mathrm{H}_2 \mathrm{O}\)), the chemical equation is used to show the reactants transforming into products: sulfur dioxide (\(\mathrm{SO}_2\)) and hydrochloric acid (\(\mathrm{HCl}\)).
  • The left side of the equation includes the reactants, which are the starting substances.
  • The right side shows the products, which are substances formed by the reaction.
  • Each substance in the reaction is represented by its chemical formula, such as \(\mathrm{SOCl}_2\), which reveals its composition.
When writing chemical equations, the reactants are separated from the products by an arrow (\( \rightarrow \)). This indicates the direction of the reaction. To fully understand a chemical equation, one must balance it, ensuring the conservation of mass and atoms.
Mole Calculations
Mole calculations are crucial to determine the amount of substances involved in chemical reactions. The mole is a unit that helps quantify substances in chemistry, making it easier to balance and calculate equations.
To find out how much of a substance interacts in a chemical reaction, we use the concept of moles along with the molar mass.
  • For example, the molar mass of \(\mathrm{SOCl}_2\) is calculated by adding the atomic masses of one sulfur (32.06 g/mol), one oxygen (16.00 g/mol), and two chlorines (2 \times 35.45 g/mol), resulting in 106.92 g/mol.
  • Using this molar mass, we can determine the moles of \(\mathrm{SOCl}_2\) by dividing the given mass by the molar mass. In our case, 35.0 grams of \(\mathrm{SOCl}_2\) is equivalent to 0.327 moles, calculated as 35.0 g / 106.92 g/mol.
Mole calculations allow chemists to predict how much of each reactant is needed or product produced by a reaction, which is vital for laboratory and industrial applications.
Reaction Balancing
Balancing chemical reactions is a foundational step in ensuring that a chemical equation adheres to the law of conservation of mass, which states that mass cannot be created or destroyed in a closed system.
In balancing reactions, we adjust the coefficients in front of chemical formulas to ensure that the number of atoms for each element is equal on both sides of the equation.
  • For instance, in the equation \(\mathrm{SOCl}_2(l) + \mathrm{H}_2\mathrm{O}(l) \rightarrow \mathrm{SO}_2(g) + \mathrm{HCl}(g)\), we initially observe that hydrogen and chlorine atoms are unbalanced.
  • To balance hydrogen and chlorine, we place a coefficient of 2 in front of \(\mathrm{HCl}(g)\), making the equation \(\mathrm{SOCl}_2(l) + \mathrm{H}_2\mathrm{O}(l) \rightarrow \mathrm{SO}_2(g) + 2\mathrm{HCl}(g)\).
Balancing ensures not only that the law of conservation is upheld, but also that stoichiometric calculations can be accurately performed. It's an essential skill for solving chemical problems and being successful in chemistry studies.

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Most popular questions from this chapter

For each of the following balanced chemical equations, calculate how many grams of the product(s) would be produced by complete reaction of 0.125 mole of the first reactant. a. \(\operatorname{AgNO}_{3}(a q)+\mathrm{LiOH}(a q) \rightarrow \mathrm{AgOH}(s)+\mathrm{LiNO}_{3}(a q)\) b. \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+3 \mathrm{CaCl}_{2}(a q) \rightarrow 2 \mathrm{AlCl}_{3}(a q)+3 \mathrm{CaSO}_{4}(s)\) c. \(\mathrm{CaCO}_{3}(s)+2 \mathrm{HCl}(a q) \rightarrow \mathrm{CaCl}_{2}(a q)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)\) d. \(2 \mathrm{C}_{4} \mathrm{H}_{10}(g)+13 \mathrm{O}_{2}(g) \rightarrow 8 \mathrm{CO}_{2}(g)+10 \mathrm{H}_{2} \mathrm{O}(g)\)

Before going to lab, a student read in his lab manual that the percent yicld for a difficult reaction to be studied was likely to be only \(40 . \%\) of the theoretical yield. The student's prelab stoichiometric calculations predict that the theoretical yicld should be \(12.5 \mathrm{~g}\). What is the student's actual yield likely to be?

For the balanced chemical equation for the combination reaction of hydrogen gas and oxygen gas $$ 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(g) $$ explain why we know that \(2 \mathrm{~g}\) of \(\mathrm{H}_{2}\) reacting with \(1 \mathrm{~g}\) of \(\mathrm{O}_{2}\) will not result in the production of \(2 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\)

Using the average atomic masses given inside the front cover of this book, calculate the mass in grams of each of the following samples. a. 2.17 moles of germanium, Ge b. \(4.24 \mathrm{mmol}\) of lead(II) chloride ( \(1 \mathrm{mmol}=1 / 1000 \mathrm{~mol}\) ) c. 0.0971 mole of ammonia, \(\mathrm{NH}_{3}\) d. \(4.26 \times 10^{3}\) moles of hexane, \(\mathrm{C}_{6} \mathrm{H}_{14}\) e. 1.71 moles of iodine monochloride, \(1 \mathrm{C}=\)

Alkali metal hydroxides are sometimes used to "serub" excess carbon dioxide from the air in closed spaces (such as submarines and spacecraft). For example, lithium hydroxide reacts with carbon dioxide according to the unbalanced chemical equation $$ \mathrm{LiOH}(s)+\mathrm{CO}_{2}(g) \rightarrow \mathrm{Li}_{2} \mathrm{CO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(g) $$ Suppose a lithium hydroxide canister contains \(155 \mathrm{~g}\) of \(\mathrm{LiOH}(s)\). What mass of \(\mathrm{CO}_{2}(g)\) will the canister be able to absorb? If it is found that after 24 hours of use the canister has absorbed \(102 \mathrm{~g}\) of carbon dioxide, what percentage of its capacity has been reached?

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