Question

Write balanced chemical equations to correspond to each of the following descriptions: (a) When sulfur trioxide gas reacts with water, a solution of sulfuric acid forms. (b) Boron sulfide, \(B_{2} S_{3}(s),\) reacts violently with water to form dissolved boric acid, \(H_{3} B O_{3},\) and hydrogen sulfide gas. (c) Phosphine, PH \(_{3}(g),\) combusts in oxygen gas to form water vapor and solid tetraphosphorus decaoxide. (d) When solid mercury(II) nitrate is heated, it decomposes to form solid mercury(II) oxide, gaseous nitrogen dioxide, and oxygen.(e) Copper metal reacts with hot concentrated sulfuric acid solution to form aqueous copper(II) sulfate, sulfur dioxide gas, and water.

Short answer

The balanced chemical equations corresponding to the given descriptions are: (a) \(SO_3(g) + H_2O(l) \rightarrow H_2SO_4(aq)\) (b) \(B_2S_3(s) + 6H_2O(l) \rightarrow 2H_3BO_3(aq) + 3H_2S(g)\) (c) \(4PH_3(g) + 5O_2(g) \rightarrow 6H_2O(g) + P_4O_{10}(s)\) (d) \(Hg(NO_3)_2(s) \rightarrow HgO(s) + 2NO_2(g) + \frac{1}{2}O_2(g)\) (e) \(Cu(s) + H_2SO_4(aq) \rightarrow CuSO_4(aq) + SO_2(g) + H_2O(l)\)

Step-by-step solution

(a) Write the unbalanced equation for sulfur trioxide reacting with water

\ For the reaction between sulfur trioxide gas and water, the reactants are sulfur trioxide (\(SO_3\)) and water (\(H_2O\)). The product formed is sulfuric acid (\(H_2SO_4\)). This gives us the unbalanced equation: \[SO_3(g) + H_2O(l) \rightarrow H_2SO_4(aq)\]

(a) Balance the equation for sulfur trioxide reacting with water

\ To balance the equation, we need to account for all the elements involved. The equation is already balanced because there is one \(S\), two \(O\), and two \(H\) atoms on each side. Therefore, the balanced equation is: \[SO_3(g) + H_2O(l) \rightarrow H_2SO_4(aq)\]

(b) Write the unbalanced equation for boron sulfide reacting with water

\ For the reaction between boron sulfide (\(B_2S_3\)) and water, the reactants are boron sulfide and water (\(H_2O\)). The products are boric acid (\(H_3BO_3\)) and hydrogen sulfide gas (\(H_2S\)). The unbalanced equation is: \[B_2S_3(s) + H_2O(l) \rightarrow H_3BO_3(aq) + H_2S(g)\]

(b) Balance the equation for boron sulfide reacting with water

\ To balance this equation, we can follow these steps: 1. Balance the \(B\) atoms: We already have two \(B\) atoms on both sides. 2. Balance the \(S\) atoms: We have 3 \(S\) atoms on the left and 1 \(S\) atom on the right. We can multiply the \(H_2S\) by 3 to balance the sulfur atoms. 3. Balance the \(H\) atoms: Now, we need to balance the hydrogen atoms. We have 6 \(H\) atoms on the right and \(2\) on the left. We can multiply the \(H_2O\) by \(3\) to balance the hydrogen atoms. The balanced equation is: \[B_2S_3(s) + 6H_2O(l) \rightarrow 2H_3BO_3(aq) + 3H_2S(g)\]

(c) Write the unbalanced equation for phosphine combustion

\ For the combustion of phosphine (\(PH_3\)), the reactants are phosphine and oxygen gas (\(O_2\)). The products are water vapor (\(H_2O\)) and tetraphosphorus decaoxide (\(P_4O_{10}\)). The unbalanced equation is: \[PH_3(g) + O_2(g) \rightarrow H_2O(g) + P_4O_{10}(s)\]

(c) Balance the equation for phosphine combustion

\ To balance this equation, we can follow these steps: 1. Balance the \(P\) atoms: We have 1 \(P\) atom on the left and 4 \(P\) atoms on the right, so multiply \(PH_3\) by 4. 2. Balance the \(H\) atoms: We now have 12 hydrogen atoms in the reactants, so multiply \(H_2O\) by 6. 3. Balance the \(O\) atoms: We have 10 \(O\) atoms on the right and 2 on the left, so multiply \(O_2\) by 5. The balanced equation is: \[4PH_3(g) + 5O_2(g) \rightarrow 6H_2O(g) + P_4O_{10}(s)\]

(d) Write the unbalanced equation for mercury(II) nitrate decomposition

\ For the decomposition of mercury(II) nitrate (\(Hg(NO_3)_2\)), the products are mercury(II) oxide (\(HgO\)), nitrogen dioxide (\(NO_2\)), and oxygen gas (\(O_2\)). The unbalanced equation is: \[Hg(NO_3)_2(s) \rightarrow HgO(s) + NO_2(g) + O_2(g)\]

(d) Balance the equation for mercury(II) nitrate decomposition

\ To balance this equation, we can follow these steps: 1. Balance the \(Hg\) atoms: We already have one \(Hg\) atom on both sides. 2. Balance the \(N\) atoms: We have 2 \(N\) atoms on the left and 1 \(N\) atom on the right, so multiply \(NO_2\) by 2. 3. Balance the \(O\) atoms: We now have 6 \(O\) atoms on the left, 4 \(O\) atoms in \(2NO_2\), and 1 oxygen atom in \(HgO\). This means we need 1 more \(O\) atom on the right, so multiply \(O_2\) by \(\frac{1}{2}\). The balanced equation is: \[Hg(NO_3)_2(s) \rightarrow HgO(s) + 2NO_2(g) + \frac{1}{2}O_2(g)\]

(e) Write the unbalanced equation for copper reacting with sulfuric acid

\ For the reaction between copper (\(Cu\)) and concentrated sulfuric acid (\(H_2SO_4\)), the products are copper(II) sulfate (\(CuSO_4\)), sulfur dioxide gas (\(SO_2\)), and water (\(H_2O\)). The unbalanced equation is: \[Cu(s) + H_2SO_4(aq) \rightarrow CuSO_4(aq) + SO_2(g) + H_2O(l)\]

(e) Balance the equation for copper reacting with sulfuric acid

\ To balance this equation, we can follow these steps: 1. Balance the \(Cu\) atoms: We already have one \(Cu\) atom on both sides. 2. Balance the \(S\) atoms: We have one \(S\) atom on both sides, so no changes are needed. 3. Balance the \(H\) atoms: We have 2 \(H\) atoms on the left and 2 on the right, so no changes are needed. 4. Balance the \(O\) atoms: We have 4 \(O\) atoms on the left and 4 on the right, so the equation is already balanced. The balanced equation is: \[Cu(s) + H_2SO_4(aq) \rightarrow CuSO_4(aq) + SO_2(g) + H_2O(l)\]

Key concepts

Chemical Reaction

When substances interact to form new products, we refer to this as a chemical reaction. The essence of a chemical reaction involves breaking old bonds and forming new ones. Each reaction is governed by certain rules that ensure the conservation of mass. This principle dictates that the number of atoms for each element involved must be the same before and after the reaction. This is where we use what is known as a balanced chemical equation.

For instance, in the reaction where sulfur trioxide gas reacts with water to produce sulfuric acid, represented by the equation:
\[SO_3(g) + H_2O(l) \rightarrow H_2SO_4(aq)\]
the equation is balanced as there is an equal number of sulfur, oxygen, and hydrogen atoms on both sides of the reaction. In essence, balanced equations are vital for accurately representing the transformation of reactants to products and for making quantitative predictions about the reactions.

Stoichiometry

The quantitative relationship between reactants and products in a chemical reaction is known as stoichiometry. It involves using the balanced chemical equations to calculate how much reactant is needed to produce a certain amount of product. In a balanced chemical equation, the coefficients represent the mole ratios of the substances involved.

Take for example the reaction of boron sulfide with water to form boric acid and hydrogen sulfide:
\[B_2S_3(s) + 6H_2O(l) \rightarrow 2H_3BO_3(aq) + 3H_2S(g)\]
The coefficients (1, 6, 2, and 3) tell us the exact ratio in which substances react and are formed. These coefficients are crucial for stoichiometry calculations as they allow us to convert between masses, volumes, and moles of reactants and products. Understanding stoichiometry is essential for performing accurate chemical calculations and for practical applications such as determining the correct proportions for a reaction mixture in industrial processes.

Combustion

A specific type of chemical reaction is combustion, which usually involves a substance (typically a hydrocarbon) reacting with oxygen to release energy in the form of heat and light, producing water and carbon dioxide in the process. However, combustion doesn't always produce carbon dioxide - it depends on the substance that is burning.

For instance, the combustion of phosphine in oxygen is a reaction that forms water and tetraphosphorus decaoxide, which is represented by the equation:\[4PH_3(g) + 5O_2(g) \rightarrow 6H_2O(g) + P_4O_{10}(s)\]
This balanced chemical equation shows the stoichiometry of the combustion process. It's important to balance combustion equations correctly to understand the amount of oxygen needed for the reaction to occur completely and to predict the quantities of the products formed. Properly understanding combustion reactions has practical applications in various fields, such as engine design, fire safety, and environmental science, where knowing the products of combustion is essential for controlling pollution.