`44` g of `C_3H_8` react with `80` g of `O_2` to form `CO_2` and `H_2O.` How many grams of `CO_2` will be formed?

First, write the balanced equation, this is simple:

`C_3H_8 + 5O_2 = 3CO_2 + 4H_2O.`

Hence each mole of propane `C_3H_8` requires `5` moles of oxygen `O_2.`

Also we need to compute molar masses: for `C_3H_8` it is `3*12+8*1 = 44 g/(mol),` for `O_2` it is `2*16 = 32 g/(mol)` and for `CO_2` it is `12+2*16 = 44 g/(mol).`

Now we can determine what initial substance will react completely: `44g` of `C_3H_8` is `1` mole and `80g` of `O_2` are `80/32 = 2.5` moles. `1` mole of `C_3H_8` requires `5` moles of `O_2,` but there are only `2.5,` so all oxygen will be used.

Finally, `5` moles of oxygen make `3` moles of `CO_2,` so `2.5` moles make `1.5` moles of `CO_2,` and they have the mass of `1.5*44 =` 66 g. This is the answer.