Chapter 15, Problem 108A

(a)

Interpretation Introduction

Interpretation: To determine the moles of oxygen required to produce 10.8 g of water.

Concept Introduction: Mole can be defined as the ratio of mass and molar mass of a particular compound.

Answer to Problem 108A

The mole of oxygen will be 0.33 mole.

Explanation of Solution

According to the balanced equation,

  ${\text{2H}}_{\text{2}}{\text{(g)+O}}_{\text{2}}\text{(g)}\to {\text{2H}}_{\text{2}}\text{O}$

1 mole of hydrogen combines with $\frac{\text{1}}{\text{2}}$ mole of oxygen to form 1 moles of water.

So, 4.5 moles of hydrogen will combine with 2.25 moles of oxygen and form 4.5 moles of water.

It is given that,

The mass of oxygen = 10.8 g

The molar mass of oxygen = 32.g

The mole of oxygen can be determined by using the formula:

  $\begin{array}{l}\begin{array}{c}\text{Number of mole = }\frac{\text{mass}}{\text{molar}\text{mass}}\\ \text{=}\frac{\text{10}\text{.8}}{\text{32}}\\ \text{=0}\text{.33}\text{mole}\end{array}\hfill \end{array}$

The moles of oxygen will be 0.33 moles.

(b)

Interpretation Introduction

Interpretation: To determine the volume of oxygen.

Concept Introduction: Mole can be defined as the ratio of mass and molar mass of a particular compound. The unit of the mole is the mole.

Answer to Problem 108A

The volume of the oxygen will be 7.37 L

Explanation of Solution

The given data:

A mole of oxygen = 0. 3375 moles

At STP,

Temperature (T) = 273 K

Pressure (P) = 1 atm

According to the ideal gas equation,

  $\text{PV=nRT}$   ......(i)

Where P is pressure, V is volume and T is temperature.

Put the given data in the above equation.

  $\begin{array}{l}\text{1 atm×V = 0}{\text{. 3375}}^{}\text{mole×0}\text{.08 × 273 K}\hfill \\ \begin{array}{c}\text{V = }\frac{\text{0}{\text{. 3375}}^{}\text{mole ×0}\text{.08 × 273 K}}{\text{1}\text{atm}}\\ \text{ = 7}\text{.37 L}\end{array}\hfill \end{array}$

Therefore, the volume of the oxygen will be 7.37L.