Textbook Problem

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Water can be synthesized according to the following unbalanced chemical equation:

${\text{H2+ O}}_{\text{2}}\to {\text{ H}}_{\text{2}}\text{O}$

a. Balance the equation.

b. How many moles of ${\text{H}}_{\text{2}}\text{O}$ can be produced if you have 2.72 mol of ${\text{O}}_{\text{2}}$ and an unlimited amount of ${\text{H}}_{\text{2}}$?

c. How many moles of ${\text{H}}_{\text{2}}$ would be necessary to produce 10.0 g of water?

d. How many grams of water would be formed by the complete reaction of 2.5 g of ${\text{H}}_{\text{2}}$?

Interpretation Introduction

Interpretation:

The chemical equation is to be balanced. Also, the number of moles of ${\text{H}}_{\text{2}}\text{O}$, number of moles of ${\text{H}}_{\text{2}}$, and the number of grams of water are to be determined.

Concept introduction:

A chemical equation is a symbolic depiction of a chemical reaction that contains symbols and formula.

In a chemical reaction, the starting substances on the left side are known as reactants, while the substances on the right side are known as products.

A conversion factor is a numerical ratio, which is used in order to convert one unit into another.

The chemical formula represents the precise number of atoms in a molecule, but not their structural arrangement.

Explanation of Solution

a) The balance equation

Oxygen atom is balanced by multiplying the coefficient $2$ with ${\text{H}}_2\text{O}$ in the product part as follows:

${\text{H}}_2+{\text{O}}_2\to 2{\text{H}}_2\text{O}$

In the next step, $\text{H}$ atom is balanced by multiplying the coefficient $2$ with ${\text{H}}_{\text{2}}$ in the reactant part as follows:

$\text{2}{\text{H}}_2+{\text{O}}_2\to 2{\text{H}}_2\text{O}$

b) Number of moles of ${\text{H}}_{\text{2}}\text{O}$ produced with $2.72{\text{mol of O}}_2$ and an unlimited amount of ${\text{H}}_{\text{2}}$.

The equivalence is obtained between moles of ${\text{O}}_2$ and moles of ${\text{H}}_{\text{2}}\text{O}$ with the help of the chemical formula as follows:

$2{\text{mol H}}_2\text{O}\equiv 1{\text{mol O}}_2$

In order to obtain the moles of ${\text{H}}_{\text{2}}\text{O}$, the conversion factor proceeds as follows:

$2.72\cancel{{\text{mol O}}_2}\times \frac{2{\text{mol H}}_2\text{O}}{1\cancel{{\text{mol O}}_2}}=5.44{\text{mol H}}_2\text{O}$

Therefore, the moles of water are $5.44\text{ mol}$.

c) Number of moles of ${\text{H}}_2$ necessary to produce $\text{10}\text{.0}\text{g of water}$

The equivalence is obtained between grams of ${\text{H}}_{\text{2}}\text{O}$ and moles of ${\text{H}}_{\text{2}}\text{O}$ with the help of the chemical formula as follows:

$1{\text{mol H}}_{\text{2}}\text{O}\equiv 18.02{\text{g H}}_{\text{2}}\text{O}$

In order to obtain the moles of ${\text{H}}_{\text{2}}\text{O}$, the conversion factor proceeds as follows:

$\text{10}\text{.0 g }\cancel{{\text{H}}_2\text{O}}\times \frac{1{\text{mol H}}_2\text{O}}{\text{18}\text{.02}\text{g }\cancel{{\text{H}}_2\text{O}}}=0.555{\text{mol H}}_2\text{O}$

Here,

$\text{Molar}{\text{mass of H}}_{\text{2}}\text{O}=18$