Question

8KCIO3 +1C12H22011

- >12CO2 + 11H20 + 8KCI
If you want the gummy bear to be the limiting reactant
for the reaction above, what is the minimum
amount of potassium chlorate that must be used to react completely with one gummy bear?
note: gummy bears are many made up of sucrose (1.1 g of sucrose in this equation)

Answer 1

3.2 g KClO3

Step-by-step explanation:

1.1 g C12H22O11 × (1 mol C12H22O11/342.3 g C12H22O11)

= 0.0032 mol C12H22O11

0.0032 mol C12H22O11 × (8 mol KClO3/1 mol C12H22O11)

= 0.026 mol KClO3

Therefore, the minimum amount of KClO3 needed is

0.026 mol KClO3 × (122.55 g KClO3/1 mol KClO3)

= 3.2 g KClO3