Question

Venus's atmosphere contains about 4.6 *10^(20) kg of carbon dioxide (CO_(2)). If all of the oxygen atoms in the CO_(2) were combined with hydrogen to make water (h_(2)o) instead of carbon dioxide, what would the total mass of water be? (Note: Hydrogen has an atomic mass of 1; carbon, 12; oxygen, 16.)

Answer 1

Answer:

$mass\text{o}f\text{w}ater=7.524\cdot10^{20}kg$

Explanations:

Given the following parameters;

Mass of CO2 = 4.6 * 10^20kg

Mass of CO2 = 4.6 * 10^23 grams

Determine the mole of CO2

$\begin{gathered} mole\text{ of CO}_2=\frac{mass}{molar\text{ mass}} \\ mole\text{ of }CO_2=\frac{4.6\times10^{23}g}{44.01g\text{/mol}} \\ mole\text{ of CO}_2=0.1045\times10^{23}mole \\ mole\text{ of CO}_2=1.045\times10^{22}moles \end{gathered}$

Since there are two atoms of oxygen in CO2, the total moles of oxygen will be expressed as:

$\begin{gathered} moles\text{ of O}_2=2\times1.045\times10^{22} \\ moles\text{ of O}_2=2.09\times10^{22}moles \end{gathered}$

The reaction between Oxygen and Hydrogen is expressed as:

$2H_2+O_2\rightarrow2H_2O$

According to stochiometry, 1mole of oxygen produces 2 moles of water, hence the moles of water required will be given as;

$\begin{gathered} mole\text{ of H}_2O=2\times2.09\times10^{22} \\ mole\text{ of H}_2O=4.18\times10^{22}moles \end{gathered}$

Determine the mass of water required:

$\begin{gathered} mass\text{ of water}=mole\times molar\text{ mass} \\ mass\text{ of water =4.18}\times10^{22}\times18 \\ mass\text{ of water=7.524}\times10^{23}g \\ mass\text{ o}f\text{ }water=7.524\times10^{20}kg \end{gathered}$

This gives the required total mass of water needed