Question

The following thermochemical equation is for the reaction of ethane(g) with oxygen(g) to form carbon dioxide(g) and water(g). 2C2H6(g) + 7O2(g)4CO2(g) + 6H2O(g) H = -2.86×103 kJ When 6.74 grams of ethane(g) react with excess oxygen(g), kJ of energy are .

Answer 1

Answer: Thus
`0.31* 10^3kJ` of energy is produced

Step-by-step explanation:

To calculate the moles :

`\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}`

`\text{Moles of} C_2H_6=(6.74g)/(30g/mol)=0.22moles`

`2C_2H_6(g)+7O_2(g)\rightarrow 4CO_2(g)+6H_2O(g)`
`\Delta H=-2.86* 10^3kJ`

According to stoichiometry

2 moles of ethane produce energy =
`2.86* 10^3kJ`

Thus 0.22moles of ethane produce energy =
`(2.86* 10^3kJ)/(2)* 0.22=0.31* 10^3kJ`

Thus
`0.31* 10^3kJ` of energy is produced