Question

Type in the box below the balanced, chemical equation that describes the complete combustion reaction of starch (C18H30O15). To receive full credit, you must include the phases of all the reactants and products as well. Hints: you should know the two products. Note that the compound being combusted contains oxygen.

Answer 1

Answer: C18H30O15 (solid) + 9 O2 (Gas) --> 18 CO2 (Gas) + 15 H20 (Liquid)

Explanation:

So we know that the products of combustion are water vapor and carbon dioxide so we can put those there under the products. Also it needs to react with fire so the other reaction is gonna be oxygen gas. So if we balance out the equations by multiplying the Carbon Dioxide to reach the necessary amount of Carbon, and water so that we can reach the required amount of Hydrogen, all we would need to do is add oxygen to the left side which we already have a molecule for that.

Answer 2

Answer:

C18H30O15(s) + 18O2(g) → 18CO2(g) + 15H2O(l)

Step-by-step explanation:

Step 1: Data given

The reactants are C18H30O15 and O2

The products of a combustion reaction are CO2 and H2O

Step 2: The unbalanced equation

C18H30O15(s) + O2(g) → CO2(g) + H2O(l)

Step 3: Balancing the equation

C18H30O15(s) + O2(g) → CO2(g) + H2O(l)

On the left side we have 18x C, on the right side we have 1x C (in CO2). To balance the amount of C on both sides, we have to multiply CO2 on the right side by 18

C18H30O15(s) + O2(g) → 18CO2(g) + H2O(l)

On the left side we have 30X H, on the right side we have 2x H (in H2O). To balance the amount of H on both sides, we have to multiply H2O on the right side by 15.

C18H30O15(s) + O2(g) → 18CO2(g) + 15H2O(l)

On the left side we have 17X O, on the right side we have 51x O (36x in CO2 and 15x in H2O). To balance the amount of O on both sides, we have to multiply O2 on the left side by 18. Now the equation is balanced.

C18H30O15(s) + 18O2(g) → 18CO2(g) + 15H2O(l)