Answer:
[tex]C_{12}H_{22}O_{11}(aq)+H_{2}O(aq)=4C_{2}H_{5}OH(aq)+4CO_{2}(g)[/tex]
Explanation:
1. First write the formula of each compound:
Sucrose : [tex]C_{12}H_{22}O_{11}[/tex]
Water : [tex]H_{2}O[/tex]
Ethyl alcohol : [tex]C_{2}H_{5}OH[/tex]
Carbon dioxide : [tex]CO_{2}[/tex]
2. Write the equation with the reactants on the left and the products on the right:
[tex]C_{12}H_{22}O_{11}+H_{2}O=C_{2}H_{5}OH+CO_{2}[/tex]
3. Put the states of matter of each compound in the reaction, as the problem says:
[tex]C_{12}H_{22}O_{11}(aq)+H_{2}O(aq)=C_{2}H_{5}OH(aq)+CO_{2}(g)[/tex]
aq = aqueous solution
g = gas
4. Balance each element.
The number of each atom must be the same on the left and on the right.
[tex]C_{12}H_{22}O_{11}(aq)+H_{2}O(aq)=4C_{2}H_{5}OH(aq)+4CO_{2}(g)[/tex]