Question

When methane ( CH4 ) burns, it reacts with oxygen gas to produce carbon dioxide and water. The unbalanced equation for this reaction isCH4(g)+O2(g)→CO2(g)+H2O(g) This type of reaction is referred to as a complete combustion reaction.What mass of water is produced from the complete combustion of 5.90×10−3 g of methane?Express your answer with the appropriate units.

Answer 1

`1.33*10^(-2)grams`

Explanations

The complete balanced equation for the given reaction is expressed as;

`CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)`

Given the following parameters

Mass of CH4 = 5.90×10^−3 g = 0.0059grams

Determine the moles of methane

`\begin{gathered} moles\text{ of CH}_4=\frac{mass}{molar\text{ mass}} \\ moles\text{ of CH}_4=(0.0059)/(16.04) \\ moles\text{ of CH}_4=0.000368moles \end{gathered}`

According to stoichimetry, 1 mole of methane produces 2 moles of water, hence the moles of water required will be:

`\begin{gathered} moles\text{ of H}_2O=(2)/(1)*0.000368 \\ moles\text{ of H}_2O=0.000736moles \end{gathered}`

Determine the mass of water produced

`\begin{gathered} Mass\text{ of H}_2O=moles* molar\text{ mass} \\ Mass\text{ of H}_2O=0.000736*18.02 \\ Mass\text{ of H}_2O=0.0133grams=1.33*10^(-2)grams \end{gathered}`

Therefore the mass of water produced from the complete combustion of 5.90×10−3 g of methane is 1.33 * 10^-2grams