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Complete and balance the following equations: $$\begin{array}{l}{\text { (a) } \operatorname{NaH}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow} \\ {\text { (b) } \mathrm{Fe}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow} \\\ {\text { (c) } \mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \longrightarrow}\end{array}$$ $$\begin{array}{l}{\text {{(d)} ) ~ } \mathrm{Na}(l)+\mathrm{H}_{2}(g) \longrightarrow} \\ {\text { (e) } \mathrm{PbO}(s)+\mathrm{H}_{2}(g) \longrightarrow}\end{array}$$

Short Answer

Expert verified
The short answer is: (a) NaH(s) + H2O(l) → NaOH(aq) + H2(g) (b) Fe(s) + H2SO4(aq) → FeSO4(aq) + H2(g) (c) H2(g) + Br2(g) → 2HBr(g) (d) 2Na(l) + H2(g) → 2NaH(s) (e) PbO(s) + H2(g) → H2O(l) + Pb(s)

Step by step solution

01

(a) Balancing NaH(s) + H2O(l) equation

: First, identify the products of the reaction: In this case, it's a reaction between a metal hydride (NaH) and water, which typically produces a metal hydroxide and hydrogen gas. Therefore, the products are NaOH and H2. Now let's balance the equation: NaH(s) + H2O(l) → NaOH(aq) + H2(g) Check if the equation is balanced: On the reactants' side, there is 1 Na, 2 H, and 1 O, and on the products' side, there is 1 Na, 2 H, and 1 O. The equation is already balanced. The balanced equation is: NaH(s) + H2O(l) → NaOH(aq) + H2(g)
02

(b) Balancing Fe(s) + H2SO4(aq) equation

: The reaction involves iron (Fe) and sulfuric acid (H2SO4), which typically forms an iron sulfate salt and hydrogen gas. The products are FeSO4 and H2. Now let's balance the equation: Fe(s) + H2SO4(aq) → FeSO4(aq) + H2(g) The equation is already balanced, with 1 Fe, 2 H, 1 S, and 4 O atoms on both sides. The balanced equation is: Fe(s) + H2SO4(aq) → FeSO4(aq) + H2(g)
03

(c) Balancing H2(g) + Br2(g) equation

: The reaction between hydrogen gas (H2) and bromine gas (Br2) forms hydrogen bromide (HBr). Now let's balance the equation: H2(g) + Br2(g) → 2HBr(g) The equation is now balanced, with 2 H and 2 Br atoms on both sides. The balanced equation is: H2(g) + Br2(g) → 2HBr(g)
04

(d) Balancing Na(l) + H2(g) equation

: The reaction between sodium (Na) and hydrogen gas (H2) forms sodium hydride (NaH). Now let's balance the equation: 2Na(l) + H2(g) → 2NaH(s) The equation is now balanced, with 2 Na and 2 H atoms on both sides. The balanced equation is: 2Na(l) + H2(g) → 2NaH(s)
05

(e) Balancing PbO(s) + H2(g) equation

: The reaction between lead oxide (PbO) and hydrogen gas (H2) forms water and lead (Pb). Now let's balance the equation: PbO(s) + H2(g) → H2O(l) + Pb(s) The equation is already balanced, with 1 Pb, 2 H, and 1 O atoms on both sides. The balanced equation is: PbO(s) + H2(g) → H2O(l) + Pb(s)

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Key Concepts

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

Chemical Reaction
A chemical reaction is a fundamental concept in chemistry that describes the process by which substances, known as reactants, are transformed into different substances called products. This transformation happens through the breaking and forming of chemical bonds. Let's consider the provided exercise where we have the reaction between sodium hydride (NaH) and water (H2O) resulting in sodium hydroxide (NaOH) and hydrogen gas (H2). It's essential to remember that a chemical reaction rearranges the atoms of the reactants to create new substances rather than creating or destroying atoms.

Understanding the nature of the reactants and predicting the products is crucial in the first step of solving a chemical equation. For instance, when a metal reacts with an acid, like iron (Fe) with sulfuric acid (H2SO4), we expect a salt and hydrogen gas to form. In this case, iron sulfate (FeSO4) and hydrogen gas are produced. Through these examples, we underscore how chemical reactions are reflected in the changes of substances and the formation of new ones.
Stoichiometry
The term stoichiometry is derived from the Greek words 'stoicheion' (element) and 'metron' (measure), which together mean the measurement of elements. Stoichiometry refers to the quantitative relationships or ratios of reactants and products in a chemical reaction. It plays a pivotal role in ensuring that equations are balanced, which means that the number of atoms for each element is the same on both sides of the equation. This balance is necessary to adhere to the law of conservation of mass.

To achieve a balanced equation, as demonstrated in the exercise, we adjust the coefficients in front of formulas. For example, balancing the reaction between hydrogen gas and bromine gas to form hydrogen bromide requires a coefficient of 2 in front of HBr to ensure 2 atoms of hydrogen and 2 atoms of bromine are present on both sides. Stoichiometry not only involves counting atoms but also involves calculating masses, volumes, and moles of the reactants and products, making it a bridge between the molecular world and the practical, measurable world.
Law of Conservation of Mass
The law of conservation of mass, established by Antoine Lavoisier in the 18th century, states that mass in an isolated system is neither created nor destroyed by chemical reactions or physical transformations. According to this law, the mass of the reactants must equal the mass of the products. When applying this law to balance chemical equations, we ensure that the quantity of each element remains the same before and after the reaction.

For example, in the reaction where sodium (Na) reacts with hydrogen gas (H2) to form sodium hydride (NaH), we must have the same amount of sodium and hydrogen atoms after the reaction as were present before it. If there were 2 atoms of sodium and 2 atoms of hydrogen on the reactants' side, there should be 2 atoms of each in the products as well. By honoring the law of conservation of mass, we solidify our understanding of physical changes in the world and sharpen our ability to predict the outcomes of chemical reactions.

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Most popular questions from this chapter

Complete and balance the following equations:$$\begin{array}{l}{\text { (a) } \mathrm{Mg}_{3} \mathrm{N}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow} \\ {\text { (b) } \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow} \\ {\text { (c) } \mathrm{N}_{2} \mathrm{O}_{5}(g)+\mathrm{H}_{2} \mathrm{O}(I) \longrightarrow}\end{array}$$ $$ \begin{array}{l}{\text {{d} ) ~ } \mathrm{NH}_{3}(a q)+\mathrm{H}^{+}(a q) \longrightarrow} \\ {\text { (e) } \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{O}_{2}(g) \longrightarrow}\end{array} $$ Which ones of these are redox reactions?

Give the chemical formula for (a) carbonic acid, (b) sodium cyanide, ( c) potassium hydrogen carbonate, (d) acetylene, (e) iron pentacarbonyl.

Which of the following statements are true? (a) Both nitrogen and phosphorus can form a pentafluo- ride compound.(b) Si can form three stable compounds containing two Si atoms each, \(\mathrm{Si}_{2} \mathrm{H}_{2}, \mathrm{Si}_{2} \mathrm{H}_{4},\) and \(\mathrm{Si}_{2} \mathrm{H}_{6}\) (c) \(\mathrm{Cl}_{2}\) is easier to oxidize than \(\mathrm{I}_{2}\) (d) At room temperature, the stable form of oxygen is \(\mathrm{O}_{2}\) , whereas that of sulfur is \(\mathrm{S}_{8} .\)

What is the anhydride for each of the following acids: (a) \(\mathrm{H}_{2} \mathrm{SO}_{4},(\mathbf{b}) \mathrm{HClO}_{3},(\mathbf{c}) \mathrm{HNO}_{2},(\mathbf{d}) \mathrm{H}_{2} \mathrm{CO}_{3},(\mathbf{e}) \mathrm{H}_{3} \mathrm{PO}_{4} ?\)

Give the chemical formula for (a) hydrocyanic acid, (b) nickel tetracarbonyl, (c) barium bicarbonate, (d) calcium acetylide, (e) potassium carbonate.

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