Question

What would be the limiting reagent if 19.2 grams of C7H16 were reacted with 120 grams of O2? C7H16+11O2=>7CO2+8H2O

Answer 1

`C_7H_16` would be the limiting reagent.

Starting from the balanced chemical equation

`C_7H_16 + 11O_2 -> 7CO_2 + 8H_2O`

we can see that we have a `1:11` mole ratio between `C_7H_16` and `O_2`; that is, for every mole of `C_7H_16` used in the reaction, `11` moles of `O_2` are required.

So, the number of `C_7H_16` moles, knowing that its molar mass is #100g/(mol), is

`n_(C_7H_16) = m_(C_7H_16)/(molarmass) = (19.2g)/(100g/(mol)) = 0.2` moles

The number of `O_2` moles is (its molar mass is `32g/(mol)`)

`n_(O_2) = m_(O_2)/(molarmass) = (120g)/(32g/(mol)) = 3.8` moles

However, the number of moles needed is

`n_(O_2) = n_(C&H_16) * 11 = 0.2 * 11 = 2.2` moles

This means that we have excess `O_2` (`3.8 - 2.2 = 1.6` moles) and `C_7H_16` is the limiting reagent.