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### H.19

Posted: Sun Oct 06, 2019 8:06 pm
Hi, I could use some help doing homework problem H.19:
Write balanced equation for the combustion of the solid C11H18N2O5 to carbon dioxide gas, liquid water, and nitrogen gas.

### Re: H.19

Posted: Sun Oct 06, 2019 8:23 pm
2 C11H18N2O5 + 17 O2 $\rightarrow$22 CO2 + 18 H2 +2 N2

### Re: H.19

Posted: Sun Oct 06, 2019 8:41 pm
Viviana Velasquez wrote:Hi, I could use some help doing homework problem H.19:
Write balanced equation for the combustion of the solid C11H18N2O5 to carbon dioxide gas, liquid water, and nitrogen gas.

Hi,

The equation for that problem is actually C14H18N2O5 to carbon dioxide gas, liquid water, and nitrogen gas. We know that combustion involves burning something in the presence of oxygen gas. So the reaction would be C14H18N2O5(s) + O2(g) -> CO2(g) + H2O(l) + N2(g). From here on we would move throughout the equations and use stoichiometric coefficients to balance the equation. Whenever I notice that elements add up to odd numbers, I try to make it even and simplify later. So instead of multiplying CO2 on the product side by 14, I would by multiply it by 28 and multiply C14H18N2O5 by 2 to get the common factor of 28. Moving on, the equation would be 2C14H18N2O5(s) + O2(g) -> 28CO2(g) +H2O(l) + N2(g). Then I would multiply N2 by 2 to equal the 4N on the reactant side. Then, I would multiply H2O by 18 to equal the 36H on the left side, making the equation 2C14H18N2O5(s) + O2(g) -> 28CO2(g) + 18H2O(l) + 2N2(g). We find that there are now 74 oxygen on the right side [(28x2) + (18)]. There are already 10 oxygen on the left side, so we only need 64 more oxygen to make both sides equal. The equation ends up becoming 2C14H18N2O5(s) + 32O2(g) -> 28CO2(g) + 18H2O(l) + 2N2(g). Reducing everything by a factor of 2 gives you C14H18N2O5(s) + 16O2(g) -> 14CO2(G) + 9H2O(l) + N2(g).