Question

Completeand balance the following equations: (a) \(\mathrm{NaH}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\) (b) \(\mathrm{Fe}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow\) (c) \(\mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \longrightarrow\) (d) \(\mathrm{Na}(l)+\mathrm{H}_{2}(g) \longrightarrow\) (e) \(\mathrm{PbO}(\mathrm{s})+\mathrm{H}_{2}(g) \longrightarrow\)

Short answer

(a) \(\mathrm{NaH}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{NaOH}(aq)+\mathrm{H}_{2}(g)\) (b) \(\mathrm{Fe}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{FeSO}_{4}(s)+\mathrm{H}_{2}(g)\) (c) \(\mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \longrightarrow 2\mathrm{HBr}(g)\) (d) \(2\mathrm{Na}(l)+\mathrm{H}_{2}(g) \longrightarrow 2\mathrm{NaH}(s)\) (e) \(\mathrm{PbO}(\mathrm{s})+\mathrm{H}_{2}(g) \longrightarrow \mathrm{Pb}(s)+\mathrm{H}_{2}\mathrm{O}(l)\)

Step-by-step solution

Identify elements

Identify all the elements on both sides of the equation: \(\mathrm{Na, H, and O}\)

Balance each element

Balance each element by adjusting the coefficients in front of the chemical formulas: 1 Na: \(\mathrm{Na}\) 2 H: \(\mathrm{H}_{2}\) 1 O: \(\mathrm{O}\) \(\mathrm{NaH}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{NaOH}(aq)+\mathrm{H}_{2}(g)\) (b) Balance the \(\mathrm{Fe}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow\)

Identify elements

Identify all the elements on both sides of the equation: \(\mathrm{Fe, H, S, and O}\)

Balance each element

Balance each element by adjusting the coefficients in front of the chemical formulas: 1 Fe: \(\mathrm{Fe}\) 2 H: \(\mathrm{H}_{2}\) 1 S: \(\mathrm{S}\) 4 O: \(\mathrm{O}_{4}\) \(\mathrm{Fe}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{FeSO}_{4}(s)+\mathrm{H}_{2}(g)\) (c) Balance the \(\mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \longrightarrow\)

Identify elements

Identify all the elements on both sides of the equation: \(\mathrm{H, and Br}\)

Balance each element

Balance each element by adjusting the coefficients in front of the chemical formulas: 2 H: \(\mathrm{H}_{2}\) 2 Br: \(\mathrm{Br}_{2}\) \(\mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \longrightarrow 2\mathrm{HBr}(g)\) (d) Balance the \(\mathrm{Na}(l)+\mathrm{H}_{2}(g) \longrightarrow\)

Identify elements

Identify all the elements on both sides of the equation: \(\mathrm{Na, and H}\)

Balance each element

Balance each element by adjusting the coefficients in front of the chemical formulas: 2 Na: \(\mathrm{Na}_{2}\) 2 H: \(\mathrm{H}_{2}\) \(2\mathrm{Na}(l)+\mathrm{H}_{2}(g) \longrightarrow 2\mathrm{NaH}(s)\) (e) Balance the \(\mathrm{PbO}(\mathrm{s})+\mathrm{H}_{2}(g) \longrightarrow\)

Identify elements

Identify all the elements on both sides of the equation: \(\mathrm{Pb, O, and H}\)

Balance each element

Balance each element by adjusting the coefficients in front of the chemical formulas: 1 Pb: \(\mathrm{Pb}\) 1 O: \(\mathrm{O}\) 2 H: \(\mathrm{H}_{2}\) \(\mathrm{PbO}(\mathrm{s})+\mathrm{H}_{2}(g) \longrightarrow \mathrm{Pb}(s)+\mathrm{H}_{2}\mathrm{O}(l)\)

Key concepts

Balancing Chemical Equations

Balancing chemical equations is essential because it shows that matter is conserved in chemical reactions. When molecules interact in a reaction, the atoms are rearranged, but the total count of each atom cannot change. Balancing an equation ensures that the number of each type of atom is the same on both sides of the equation.

Understanding how to balance a chemical equation starts with identifying each element involved in the reaction. Then, you adjust the numerical coefficients in front of the formulas until the number of atoms for each element is equal on both sides. This often involves some trial and error, and practice makes this process easier.

For instance, consider the reaction between sodium hydride (\(\text{NaH}\)) and water (\(\text{H}_2\text{O}\)). The balanced equation is:

• \(\text{NaH}(s) + \text{H}_2\text{O}(l) \rightarrow \text{NaOH}(aq) + \text{H}_2(g)\)

Notice how this equation is balanced with one sodium (\(\text{Na}\)), two hydrogen (\(\text{H}\)), and one oxygen (\(\text{O}\)) on each side. Balancing ensures that the mass of reactants equals the mass of products, complying with the law of conservation of mass.

Chemical Reactions

Chemical reactions are processes where substances, known as reactants, are transformed into different substances, called products. This transformation occurs through the breaking and forming of chemical bonds. Reactants interact at the molecular level, and through various steps, product molecules are generated.

A chemical reaction is usually denoted by a chemical equation. For example, when iron reacts with sulfuric acid, the equation is:

• \(\text{Fe}(s) + \text{H}_2\text{SO}_4(aq) \rightarrow \text{FeSO}_4(s) + \text{H}_2(g)\)

This equation shows that solid iron (\(\text{Fe}\)) reacts with aqueous sulfuric acid (\(\text{H}_2\text{SO}_4\)) to form solid iron(II) sulfate and hydrogen gas.

Every chemical reaction follows a certain pathway which involves energy changes. Reactions often absorb or release energy, which can be observed as changes in temperature or the emission of light. Understanding these reactions allows chemists to harness or counteract these energy changes in various applications.

Stoichiometry

Stoichiometry involves the quantitative relationships between reactants and products in a chemical reaction. It is based on the balanced chemical equation's coefficients, which indicate the ratio in which reactants combine and products form. Understanding this concept is crucial for calculations in chemistry, like determining how much product can be made or how much reactant is needed.

When working with stoichiometry, it is essential to convert quantities into moles, the standard unit for chemical reactions. Consider the reaction between hydrogen and bromine:

• \(\text{H}_2(g) + \text{Br}_2(g) \rightarrow 2\text{HBr}(g)\)

This balanced equation shows that one mole of hydrogen reacts with one mole of bromine to produce two moles of hydrogen bromide.

The coefficients tell us the proportions:

• 1 mole \(\text{H}_2\)
• 1 mole \(\text{Br}_2\)
• 2 moles \(\text{HBr}\)

Having these ratios allows for the calculation of unknown quantities in reactions, which is a significant part of stoichiometric calculations in chemistry.