Chapter 2.4, Problem 38LC

Interpretation Introduction

Interpretation: The amount of water formed by reacting 4.8 grams of hydrogen with 38.4 grams of oxygen should be calculated.

Concept Introduction: In a chemical change, the chemical bonds present in reactants are broken to form elements that are then rearranged to form new chemical bonds resulting in the formation of products that possess different compositions than the reactants.

Answer to Problem 38LC

  ${\text{mass of product (H}}_{\text{2}}\text{O) = 43}\text{.2 g}$

Explanation of Solution

The balanced chemical reaction between hydrogen and oxygen to form water is:

  ${\text{2H}}_{\text{2}}{\text{ + O}}_{\text{2}}\to {\text{ 2H}}_{\text{2}}\text{O}$

From the above-balanced reaction, it can be observed that 2 moles of hydrogen react with one mole of oxygen to form 2 moles of water.

Calculating the number of moles of hydrogen and oxygen using the formula:

  $\text{number of mole = }\frac{\text{given mass of the substance}}{\text{Molar mass of the substance}}$

The number of moles of hydrogen:

  $\begin{array}{l}{\text{n}}_{{\text{H}}_{\text{2}}}\text{ = }\frac{\text{4}\text{.8 g}}{\text{2 g/mol}}\\ {\text{n}}_{{\text{H}}_{\text{2}}}\text{ = 2}\text{.4 mol}\end{array}$

The number of moles of oxygen:

  $\begin{array}{l}{\text{n}}_{{\text{O}}_{\text{2}}}\text{ = }\frac{\text{38}\text{.4 g}}{\text{32 g/mol}}\\ {\text{n}}_{{\text{O}}_{\text{2}}}\text{ = 1}\text{.2 mol}\end{array}$

Since, the molar ratio of hydrogen and oxygen is the same as of balanced reaction that is 2:1 so, according to the law of conservation of mass which states that mass is conserved in any chemical or physical change:

The mass of reactants is always equal to the mass of products during any chemical reaction. Thus,

  $\begin{array}{l}{\text{mass of reactant (H}}_{\text{2}}{\text{+ O}}_{\text{2}}{\text{) = mass of product (H}}_{\text{2}}\text{O)}\\ \text{4}\text{.8 g + 38}{\text{.4 g = mass of product (H}}_{\text{2}}\text{O)}\\ {\text{mass of product (H}}_{\text{2}}\text{O) = 43}\text{.2 g}\end{array}$