Question

82. When 233.1 g of ethylene (C2H4) burns in oxygen to give carbon dioxide and water, how many grams of CO2 are formed? A) 731.4 g B) 365.7 g C) 182.9 g D) 8.31 g E) 299.4 g

Answer 1

`"Option A"`

Explanation:

We need a stoichiometric equation to represent the combustion of ethylene....

`H_2C=CH_2(g) + 3O_2(g) rarr 2CO_2(g) + 2H_2O(l) + Delta`

Charge is balanced, and mass is balanced so this is kosher...

`"Moles of ethylene"=(233.1*g)/(28.05*g*mol^-1)=8.31*mol`

And thus a mass with respect to carbon dioxide of...

`2xx8.31*molxx44.01*g*mol^-1=??*g`

And given the equation, which you should be able to reproduce standing on your head, there are `2xx8.31*mol` evolved with respect to carbon dioxide; take that atmosphere!