Question

Question #d758e

Answer 1

Your reaction will produce 14 g of carbon dioxide.

Start with the balanced chemical equation for the combustion of ethyne

`2C_2H_2 + color(red)(5)O_2 -> color(blue)(4)CO_2 + 2H_2O`

Notice that you have a `color(red)(5):color(blue)(4)` mole ratio between oxygen gas and carbon dioxide. This means that, regardless of how many moles of oxygen react, you'll always produce 0.8 times (4/5) fewer moles of carbon dioxide.

Use oxygen gas' molar mass to determine how many moles of oxygen react

`13cancel("g") * ("1 mole "O_2)/(32.0cancel("g")) = "0.406 moles"` `O_2`

Since ethyne is in excess, all the available moles of oxygen will react. This means that the number of moles of carbon dioxide the reaction will produce will be equal to

`0.406cancel("moles"O_2) * (color(blue)(4)"moles"CO_2)/(color(red)(5)cancel("moles"O_2)) = "0.325 moles"` `CO_2`

Use carbon dioxide's molar mass to see how many grams you'd get

`0.325cancel("moles") * "44.01 g"/(1cancel("mole"CO_2)) = "14.29 g"` `CO_2`

Rounded to two sig figs, the number of sig figs you gave for the mass of oxygen, the answer will be

`m_(CO_2) = color(green)("14 g")`