Question

In a butane lighter, 9.7 g of butane combine with with 34.7 g of oxygen to form g g carbon dioxide and 29.3 how many grams of water?

Answer 1

Approx. `15*g`......

Explanation:

We need a stoichiometric equation........

`C_4H_10(g) + 13/2O_2(g) rarr 4CO_2(g) + 5H_2O(l)`

And thus `"moles of butane"=(9.7*g)/(58.12*g*mol^-1)=0.167*mol`.

And so upon complete combustion we gets......

`0.167*molxx5xx18.011*g*mol^-1=15.0*g` water.......

We could have avoided this rigmarole by noting that mass is always with conserved in a chemical reaction. By the problem's specification, we had `(34.7+9.7)*g=44.4*g` of reactant; we got a mass `29.3*g` carbon dioxide, and thus there was a balance of `44.4-29.3*g=15.0*g` as required........WHICH MUST REPRESENT THE MASS OF WATER...........