Question

methane is produced and burned at a rate of approximately 4.49 x 10^6 grams per day. In the combustion process methane reacts with oxygen and produces water and carbon dioxide according to the balanced equation CH4 + 2 O2 -----> CO2 + 2 H20calculate the mass in grams of water produced by this reaction in one day.

Answer 1

1.01 * 10^7 grams/day

Explanations:

Given the reaction produced during the combustion of methane expressed as:

`CH_4+2O_2\rightarrow CO_2+2H_2O`

Calculate the moles of methane:

`\begin{gathered} Moles=\frac{Mass}{molar\text{ mass}} \\ Moles\text{ of CH}_4=(4.49*10^6)/(16.04) \\ moles\text{ of CH}_4=2.799*10^5moles \end{gathered}`

According to stochiometry, 1 mole of methane produces 2 moles of water. The moles of water produced at the end of the reaction will be:

`\begin{gathered} moles\text{ of H}_2O=2_*2.799*10^5 \\ moles\text{ of }H_2O=5.598*10^5moles \end{gathered}`

Determine the mass of water produced in a day

`\begin{gathered} Mass=moles* molar\text{ mass} \\ Mass=5.598*10^5*18.02 \\ Mass\text{ of water = 100.88}*10^5grams \\ Mass\text{ of water}=1.01*10^7grams\text{/day} \end{gathered}`

Hence the mass in grams of water produced by this reaction in one day is approximately 1.01 * 10^7 grams