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?

1 Answer
anor277
Jul 14, 2017

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...........

Related questions
See all questions in Chemical Reactions and Equations
Impact of this question
10622 views around the world
You can reuse this answer
Creative Commons License