Question

Answer the following questions by referring to the equation below:?

`2text(C)_6text(H)_10 + 17text(O)_2 -> 12text(CO)_2 + 10text(H)_2text(O)`

a) If `35g` of `"C"_6"H"_10` and `45g` of oxygen are reacted:
i) Which substance is the limiting reagent?
ii) How many grams of carbon dioxide will be formed?

b) If 35 grams of carbon dioxide are formed, calculate the percent yield of this reaction.

Answer 1

See Below

Explanation:

To figure out limiting reagent, I usually just react one species and see how much product can be formed....and then react the other species and see how much product can be formed (assuming excess of the other). Whichever makes the least is the one you'll run out of, and is your limiting reagent. However much you can make, too, is your theoretical yield.

`C_6H_10` (assuming excess oxygen)
35g`C_6H_10("1mol"/ "82.1g")` = 0.426 moles `C_6H_10`
`C_6H_10` can make how much `CO_2`?
0.426 moles `C_6H_10((12mol CO_2)/(2mol C_6H_10))` = 2.556 moles `CO_2`

`O_2` (assuming excess `C_6H_10`)
46g`O_2("1mol"/ "32g")` = 1.4375 moles `O_2`
`O_2` can make how much `CO_2`?
1.4375 moles `O_2((12mol CO_2)/(17mol O_2))` = 1.015 moles `CO_2`

`O_2` can make the least `CO_2`, so it is the limiting reagent.
The maximum amount of `CO_2` you can make is 1.015 moles

Theoretical Yield `CO_2`
1.015 moles `CO_2("44g"/"1 mol")` = 44.65g `CO_2`

Percent Yield
Percent yield is `("what you got"/"what you should have got")xx100`
Percent Yield = `("35g"/"44.65g")xx100` = 78.4%