How many moles of `CO_2` are produced from `3` kg of `C_6H_(14)` completely burned?

`C_6H_(14)` is a hydrocarbon called hexane. When burned, it produces carbon dioxide `CO_2` and water `H_2O.`

To answer the question we need to write the balanced chemical reaction. It is

`2C_6H_14 + 19O_2 = 12CO_2 + 14H_2O.`

To obtain these coefficients, we first look at carbon, `6` at the left, and give `CO_2` the coefficient `6.` Then count `H,` `14` at the left, so `H_2O` must have the coefficient `7.` But then there will be even quantity of `O` at the left and odd at the right. This requires to change `7` at `H_2O` to `14,` `1` at `C_6H_14` to `2` and `6` at `CO_2` to `12.` Then the coefficient at `O_2` becomes `19.`

Now we see that `2` moles of `C_6H_14` make `12` moles of `CO_2,` so `1` mole makes `6.` The molar mass of hexane is `6*12+14*1=86 g/(mol).`

Finally, `3 kg` of hexane are `(3000 g)/(86 g/(mol)) approx 34.9` moles, so there will be about `34.9*6 approx 209` moles of `CO_2.`